Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
[tex]\sf\underline{\underline{Question:}}[/tex]
If sum of first 6 digits of AP is 36 and that of the first 16 terms is 255,then find the sum of first ten terms.
$\sf\underline{\underline{Solution:}}$
$\sf\bold\purple{||100||}$$\space$
$\sf\underline\bold\red{||Step-by-Step||}$
$\sf\bold{Given:}$
$\sf\bold{S6=36}$ $\sf\bold{S16=255}$$\space$
$\sf\bold{To\:find:}$
$\sf\bold{The \: sum\:of\:the\:first\:ten\:numbers}$$\space$
$\sf\bold{Formula\:we\:are\:using:}$
$\implies$ $\sf{ Sn=}$ $\sf\dfrac{N}{2}$ $\sf\small{[2a+(n-1)d]}$
$\space$
$\sf\bold{Substituting\:the\:values:}$
→ $\sf{S6=}$ $\sf\dfrac{6}{2}$ $\sf\small{[2a+(6-1)d]}$
→ $\sf{36 = 3[2a+(6-1)d]}$
→$\sf{12=[2a+5d]}$ $\sf\bold\purple{(First \: equation)}$
$\space$
$\sf\bold{Again,Substituting \: the\:values:}$
→ $\sf{S16}$ $\sf\dfrac{16}{2}$ $\sf\small{[2a+(16-1)d]}$
→ $\sf{255=8[2a + (16-1)d]}$
:: $\sf\dfrac{255}{8}$ $\sf\small{=31.89=32}$
→ $\sf{32=[2a+15d]}$ $\sf\bold\purple{(Second\:equation)}$
$\space$
$\sf\bold{Now,Solve \: equation \: 1 \:and \:2:}$
→ $\sf{10=20}$
→ $\sf{d=}$ $\sf\dfrac{20}{10}$ $\sf{=2}$
$\space$
$\sf\bold{Putting \: d=2\: in \:equation - 1:}$
→ $\sf{12=2a+5\times 2}$
→ $\sf{a = 1}$
$\space$
$\sf\bold{All\:of\:the\:above\:eq\: In \: S10\:formula:}$
$\mapsto$ $\sf{S10=}$ $\sf\dfrac{10}{2}$ $\sf\small{[2\times1+(10-1)d]}$
$\mapsto$ $\sf{5(2\times1+9\times2)}$
$\mapsto$ $\sf\bold\purple{5(2+18)=100}$
$\space$
$\sf\small\red{||Hence , the \: sum\: of \: the \: first\:10\: terms\: is\:100||}$
_____________________________
Question 6 a-c if plz show ALL STEPS like LITERALLY EVERYTHING
9514 1404 393
Answer:
a) quadrant III
b) 4π/3, -2π/3
c) (1/2, (√3)/2)
d) ((√3)/2, -1/2
Step-by-step explanation:
a) The attachment shows the point P and the numbering of the quadrants (in Roman numerals). Point P lies in quadrant III.
__
b) Measured counterclockwise, the angle to point P is 240° or 4π/3 radians. Measured clockwise, the angle is -120° or -2π/3 radians. In the diagram, these are shown in green and purple, respectively.
__
c) Adding π/2 to the angle 4π/3 or -2π/3 brings it to 11π/6, or -π/6. This point is marked as P' (blue) on the diagram. The coordinate transformation for π/2 radians CCW rotation is ...
(x, y) ⇒ (-y, x)
P(-1/2, -√3/2) ⇒ P'(√3/2, -1/2)
In terms of trig functions, the coordinates of the rotated point are ...
P'(cos(-π/6), sin(-π/6)) = P'(√3/2, -1/2)
__
d) Adding or subtracting π radians to/from the angle moves it directly opposite the origin. Both coordinates change sign. This point is P'' (red) on the diagram.
[tex]\int\limits^a_b {(1-x^{2} )^{3/2} } \, dx[/tex]
First integrate the indefinite integral,
[tex]\int(1-x^2)^{3/2}dx[/tex]
Let [tex]x=\sin(u)[/tex] which will make [tex]dx=\cos(u)du[/tex].
Then
[tex](1-x^2)^{3/2}=(1-\sin^2(u))^{3/2}=\cos^3(u)[/tex] which makes [tex]u=\arcsin(x)[/tex] and our integral is reshaped,
[tex]\int\cos^4(u)du[/tex]
Use reduction formula,
[tex]\int\cos^m(u)du=\frac{1}{m}\sin(u)\cos^{m-1}(u)+\frac{m-1}{m}\int\cos^{m-2}(u)du[/tex]
to get,
[tex]\int\cos^4(u)du=\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{4}\int\cos^2(u)du[/tex]
Notice that,
[tex]\cos^2(u)=\frac{1}{2}(\cos(2u)+1)[/tex]
Then integrate the obtained sum,
[tex]\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\int\cos(2u)du+\frac{3}{8}\int1du[/tex]
Now introduce [tex]s=2u\implies ds=2du[/tex] and substitute and integrate to get,
[tex]\frac{3\sin(s)}{16}+\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\int1du[/tex]
[tex]\frac{3\sin(s)}{16}+\frac{3u}{4}+\frac{1}{4}\sin(u)\cos^3(u)+C[/tex]
Substitute 2u back for s,
[tex]\frac{3u}{8}+\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\sin(u)\cos(u)+C[/tex]
Substitute [tex]\sin^{-1}[/tex] for u and simplify with [tex]\cos(\arcsin(x))=\sqrt{1-x^2}[/tex] to get the result,
[tex]\boxed{\frac{1}{8}(x\sqrt{1-x^2}(5-2x^2)+3\arcsin(x))+C}[/tex]
Let [tex]F(x)=\frac{1}{8}(x\sqrt{1-x^2}(5-2x^2)+3\arcsin(x))+C[/tex]
Apply definite integral evaluation from b to a, [tex]F(x)\Big|_b^a[/tex],
[tex]F(x)\Big|_b^a=F(a)-F(b)=\boxed{\frac{1}{8}(a\sqrt{1-a^2}(5-2a^2)+3\arcsin(a))-\frac{1}{8}(b\sqrt{1-b^2}(5-2b^2)+3\arcsin(b))}[/tex]
Hope this helps :)
Answer:[tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}[/tex]General Formulas and Concepts:
Pre-Calculus
Trigonometric IdentitiesCalculus
Differentiation
DerivativesDerivative NotationIntegration
IntegralsDefinite/Indefinite IntegralsIntegration Constant CIntegration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
U-Substitution
Trigonometric SubstitutionReduction Formula: [tex]\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx[/tex]
Step 2: Integrate Pt. 1
Identify variables for u-substitution (trigonometric substitution).
Set u: [tex]\displaystyle x = sin(u)[/tex][u] Differentiate [Trigonometric Differentiation]: [tex]\displaystyle dx = cos(u) \ du[/tex]Rewrite u: [tex]\displaystyle u = arcsin(x)[/tex]Step 3: Integrate Pt. 2
[Integral] Trigonometric Substitution: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du[/tex][Integrand] Rewrite: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du[/tex][Integrand] Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du[/tex][Integral] Reduction Formula: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b[/tex][Integral] Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du[/tex][Integral] Reduction Formula: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg][/tex][Integral] Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg][/tex][Integral] Reverse Power Rule: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg][/tex]Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b[/tex]Back-Substitute: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b[/tex]Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b[/tex]Rewrite: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
What is the equation of a circle with a center at (4, -9) and a radius of 5?
Answer:
(x - 4)² + (y + 9)² = 25
Step-by-step explanation:
The equation of a circle is written as seen below.
(x – h)² + (y – k)² = r²
Where (h,k) represents the center of the circle and r represents the radius
We want to find the equation of a circle that has a center at (4,-9) and a radius of 5.
We know that (h,k) represents the center so h = 4 and k = -9
We also know that r represents the radius so r = 5
Now to find the equation of this specific circle we simply plug in these values into the equation of a circle formula
Equation: (x – h)² + (y – k)² = r²
h = 4, k = -9 and r = 5
Plug in values
(x - 4)² + (y - (-9))² = 5²
5² = 25
The two negative signs in front of the 9 cancel out and it changes to + 9
The equation of a circle with a center at (4,-9) and a radius of 5 is
(x - 4)² + (y + 9)² = 25
The median age (in years) of the U.S. population over the decades from 1960 through 2010 is given by
f(t) = −0.2176t3 + 1.962t2 − 2.833t + 29.4 (0 ≤ t ≤ 5)
where t is measured in decades, with t = 0 corresponding to 1960.
(a) What was the median age of the population in the year 1970?
(b) At what rate was the median age of the population changing in the year 1970?
(c) Calculate f ''(1).
Considering the given function, we have that:
a) 28.31 years.
b) 0.3382 years a decade.
c) 2.6184.
What is the function?The median age of the U.S. population in t decades after 1960 is:
f(t) = -0.2176t³ + 1.962t² - 2.833t + 29.4.
1970 is one decade after 1960, hence the median was:
f(1) = -0.2176 x 1³ + 1.962 x 1² - 2.833 x 1 + 29.4 = 28.31 years.
The rate of change was is the derivative when t = 1, hence:
f'(t) = -0.6528t² + 3.924t - 2.933
f'(1) = -0.6528 x 1² + 3.924 x 1 - 2.933 = 0.3382 years a decade.
The second derivative is:
f''(t) = -1.3056t + 3.924
Hence:
f''(1) = -1.3056 x 1 + 3.924 = 2.6184.
More can be learned about functions at https://brainly.com/question/25537936
#SPJ1
Order the expressions from least to greatest.
Anwser
4 then 5 then 6
Answer:
This the right order:
4^2+2^2 = 20
5^2= 25
6^2-6 = 30
Miya is picking up two friends to go the beach. She drives from her house to
pick up Drea, then she drives to pick up Francine, and then they go to the
beach to play volleyball. What is the total distance of the trip?
The grid below shows the coordinates of their houses on a map. All distances
are in miles.
Answer:
Option (C)
Step-by-step explanation:
Coordinates of the point representing the location of Miya → (1, 10)
Coordinates of the point representing the location of Drea → (13, 1)
Coordinates of the point representing the location of Francine → (16, 1)
Coordinates of the point representing the location of Beach → (19, 4)
Distance between Miya and Drea = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
= [tex]\sqrt{(13-1)^2+(1-10)^2}[/tex]
= [tex]\sqrt{144+81}[/tex]
= 15 miles
Distance between Drea and Francine = [tex]\sqrt{(16-13)^2+(1-1)^2}[/tex]
= 3 miles
Distance between Francine and Beach = [tex]\sqrt{(19-16)^2+(4-1)^2}[/tex]
= [tex]\sqrt{9+9}[/tex]
≈ 4.2 miles
Total distance between Miya and the beach = 15 + 3 + 4.2
= 22.2 miles
Option (C) is the answer.
Which matrix equation represents the system of equations?
Answer:
B. [tex]\left[\begin{array}{ccc}-1&2\\0&1\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] =\left[\begin{array}{ccc}0\\-2\\\end{array}\right][/tex]
Step-by-step explanation:
Given the systems of equations
-x + 2y = 0
y = -2
This can also be written as:
-x + 2y = 0
0x + y = -2
We are to write in this form AX = b
A is a 2by2 matrix with coefficients of x nd y
X is a column matrix containing the unknown
b is a column matrix with the values at the right hand sides (0 and -2)
Writing in matrix form;
[tex]\left[\begin{array}{ccc}-1&2\\0&1\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] =\left[\begin{array}{ccc}0\\-2\\\end{array}\right][/tex]
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y2 = 2x, x = 2y;
about the y-axis
b) Sketch the region
c) Sketch the solid, and a typical disk or washer.
Answer:
V = 34,13*π cubic units
Step-by-step explanation: See Annex
We find the common points of the two curves, solving the system of equations:
y² = 2*x x = 2*y ⇒ y = x/2
(x/2)² = 2*x
x²/4 = 2*x
x = 2*4 x = 8 and y = 8/2 y = 4
Then point P ( 8 ; 4 )
The other point Q is Q ( 0; 0)
From these two points, we get the integration limits for dy ( 0 , 4 )are the integration limits.
Now with the help of geogebra we have: In the annex segment ABCD is dy then
V = π *∫₀⁴ (R² - r² ) *dy = π *∫₀⁴ (2*y)² - (y²/2)² dy = π * ∫₀⁴ [(4y²) - y⁴/4 ] dy
V = π * [(4/3)y³ - (1/20)y⁵] |₀⁴
V = π * [ (4/3)*4³ - 0 - 1/20)*1024 + 0 )
V = π * [256/3 - 51,20]
V = 34,13*π cubic units
The sum of two numbers is -17. Their difference is 41. Find the numbers
Answer:
x = 12
y = -29
Step-by-step explanation:
Our given equations: x + y = -17 and x - y = 41
Solve for x and substitute.
x = -17 - y
(-17 - y) - y = 41
-17 - 2y = 41
2y = -58
y = -29
Solve for x using y
x + (-29) = -17
x = 12
Tyra has recently inherited $5400, which she wants to deposit into an IRA account. She has determined that her two best bets are an account that compounds semi-
annually at an annual rate of 3.1 % (Account 1) and an account that compounds continuously at an annual rate of 4 % (Account 2).
Step 2 of 2: How much would Tyra's balance be from Account 2 over 3.7 years? Round to two decimal places.
The focus here is the use of "Compounding interest rate" and these entails addition of interest to the principal sum of the deposit.
Tyra will definitely prefer the Account 2 over the Account 1 Tyra balance from account 2 over 3.7 years is $6,261.37
The below calculation is to derive maturity value when annual rate of 3.1% is applied.
Principal = $5,400
Annual rate = 3.1% semi-annually for 1 years
A = P(1+r/m)^n*t where n=1, t=2
A = 5,400*(1 + 0.031/2)^1*2
A = 5,400*(1.0155)^2
A = 5,400*1.03124025
A = 5568.69735
A = $5,568.70.
In conclusion, the accrued value she will get after one years for this account is $5,568.70,
- The below calculation is to derive maturity value when the amount compounds continuously at an annual rate of 4%
Principal = $5,400
Annual rate = 4% continuously
A = P.e^rt where n=1
A = 5,400 * e^(0.04*1)
A = 5,400 * 1.04081077419
A = 5620.378180626
A = 5620.378180626
A = $5,620.39.
In conclusion, the accrued value she will get after one years for this account is $5,620.39.
Referring to how much would Tyra's balance be from Account 2 over 3.7 years. It is calculated as follows:
Annual rate = 4% continuously
A = P.e^rt where n=3.7
A = 5,400 * e^(0.04*3.7)
A = 5,400 * e^0.148
A = 5,400 * 1.15951289636
A = 6261.369640344
A = $6,261.37
Therefore, the accrued value she will get after 3.7 years for this account is $6,261.37
Learn more about Annual rate here
brainly.com/question/14170671
Answer this question that is given
Answer:
See explanation
Step-by-step explanation:
2) (10+4) x 2 = 28
3) (13 + 6) x 2 = 38
4) (8+4) x 2 = 24
5) (11+8) x 2 = 38
Answered by Gauthmath
Figure A is a scale image of figure B. Figure A maps to figure B with a scale factor of 2/3. What is the value of x?
9514 1404 393
Answer:
x = 7
Step-by-step explanation:
10.5 maps to x with a scale factor of 2/3:
x = 10.5 × 2/3
x = 7
(sin 10° + cos 10º)2 - 2 sin 80°cos 80°
Answer
the simplied form of the given question is 1
Vietnamese: Phát biểu định nghĩa hàm số liên tục. Khảo sát tính liên tục của hàm số sau:
Answer:
litrally I don't understand what you are telling
An observer, who is standing 47 m from a building, measures the angle of elevation of the top of the building as 30˚. If the observer’s eye is 167 cm from the ground, what is the exact height of the building?
9514 1404 393
Answer:
28.81 m
Step-by-step explanation:
The tangent relation can help find the height of the building above the observer's eye.
Tan = Opposite/Adjacent
Opposite = Adjacent·Tan
above eye height = (47 m)(tan 30°) ≈ 27.14 m
Adding this to the eye height gives the height of the building above the ground where the observer is standing.
27.14 m + 1.67 m = 28.81 m
The height of the building to the nearest centimeter is 28.81 meters.
which of the following is the formula in solving for the area of a circle?
A.A=2πr
B.A=πr²
C.A=πd
D.A=2πr²
The area of circle is πr²
Answer:
πr²
Step-by-step explanation:
The answer is πr² where,
π = pi, 3.14...
r = radius
This is the most common way of solving for the area of the circle.
Teddy wants to taste all of the flavors of ice cream at the mall, one by one. Tasting any one flavor will change the way the next flavor taste after it. The flavors are chocolate, vanilla, strawberry, birthday cake, Rocky Road, and butter pecan. In how many ways can he taste the ice cream.
A. 30
B.120
C. 360
D.720
Answer: (d)
Step-by-step explanation:
Given
There are six flavors of ice-cream that is chocolate, vanilla, strawberry, birthday cake, rocky road, and butter pecan
First ice-cream can be tasted in 6 different ways
Second can be in 5 ways
similarly, remaining in 4, 3, 2 and 1 ways
Total no of ways are [tex]6\times5\times 4\times 3\times 2\times 1=720\ \text{ways}[/tex]
Option (d) is correct.
If a ∥ b and b ⊥ y, then _____
Answer:
a ⊥ y
Step-by-step explanation:
since b is parallel to a & perpendicular to y , line y will eventually cut across line a at a 90 degree angle as well
Answer:
a ⊥ y
Step-by-step explanation:
Look at the image given below.
2kg of chicken
61.5 g left
How many kg of chicken were eaten
Answer:
1.9385 kilograms were eaten
1.9385 kg
Step-by-step explanation:
because 2 kg=2000 g
Subtracting 2000
- 61.5
1938.5
Converting 1938.5 in kg is 1.9385 kg
Can someone please help!!! Sec. 12.7 #78
Which of the following integrals represents the volume of the solid obtained by rotating the region bounded by the curves y = (x - 2)^4 and 8x - y =16 about the line x= 10?
A. Pi integral^4_2 {[10 - (1/8 y + 2)^2] - [10 - (2 + ^4 squareroot y)^2]} dy
B. Pi integral^16_0 {[10 - (1/8 y + 2)] - [10 - (2 + ^4 Squareroot)]}^2 dy
C. Pi integral^4_2 {[10 - (1/8 y + 2)] - [10 - 2 + ^4 squareroot y)]}^2 dy
D.Pi integral^16_0 {[10 - (1/8 y + 2)]^2 - [10 - 2 + ^4 squareroot y)]^2} dy
E. Pi integral^16_0 {[10 - (1/8 y + 2)^2] - [10 - 2 + ^4 squareroot y)^2]} dy
F. Pi integral^4_2 {[10 - (1/8 y + 2)]^2 - [10 - 2 + ^4 squareroot y)]^2} dy
Answer:
[tex]\displaystyle V = \pi \int _0^{16}\left[10-\left(\frac{1}{8}y-2\right)\right] ^2 - \left[10 - \left(2+y^{{}^{1}\!/\!{}_{4}}\right)\right]^2\, dy[/tex]
Step-by-step explanation:
We want to find the volume of the solid obtained by rotating the region between the two curves:
[tex]y=(x-2)^4\text{ and } 8x-y=16[/tex]
About the line x = 16.
Since our axis of revolution is vertical, we can use the washer method in terms of y.
[tex]\displaystyle V = \pi \int _c^d[R(y)]^2 -[r(y)}]^2\, dy[/tex]
Where R(y) is the outer radius and r(y) is the inner radius.
First, solve each equation in terms of y:
[tex]\displaystyle x_1 = \frac{1}{8}y+2\text{ and } x_2 = y^{{}^{1}\! /\! {}_{4}}+2[/tex]
From the diagram below, we can see that the outer radius R(y) is (10 - x₁) and that the inner radius r(y) is (10 - x₂). The limits of integration will be from y = 0 to y = 16. Substitute:
[tex]\displaystyle V = \pi \int_0^{16}\left[\underbrace{10-\left(\frac{1}{8}y+2\right)}_{R(y)}\right]^2 - \left[\underbrace{10-\left(y^{{}^{1}\!/\!{}_{4}}+2\right)}_{r(y)}\right]^2\, dy[/tex]
Thus, our volume is:
[tex]\displaystyle V = \pi \int _0^{16}\left[10-\left(\frac{1}{8}y-2\right)\right] ^2 - \left[10 - \left(2+y^{{}^{1}\!/\!{}_{4}}\right)\right]^2\, dy[/tex]
*I labeled the diagram incorrectly. Let R(x) be R(y) and r(x) be r(y).
(a+b)2= c+d
answer
answer
Answer:
a+b=c+d/2
i cant understand what answer you want
A nurse works for a temporary nursing agency. The starting hourly wages for the six different work locations are $12.50, $11.75, $9.84, $17.67, $13.88, and $12.98. As the payroll clerk for the temporary nursing agency, find the median starting hourly wage.
please help me with this
Answer:
2x-30 + x + 40 = 180
3x + 10 = 180
3x = 170
x = 56 [tex]\frac{2}{3}[/tex]
Step-by-step explanation:
After completing the fraction division 5 / 5/3, Miko used the multiplication shown to check her work.
3 x 5/3 = 3/1 x 5/3 = 15/3 or 5
Answer:
its the same above
Step-by-step explanation:
Derive the equation of the parabola with a focus at (2,4) and a directrix of y=8
Answer:
The equation of the parabola with a focus at (2,4) and a directrix of y=8 is,
48-8y=(x-2)²
WILL MARK BRAINLIEST PLEASE SHOW WORK :)
Answer:
(1). A = 18 cm² ; (2). TR = 18 units
Step-by-step explanation:
A hot air balloon is rising vertically with a velocity of 12.0 feet per second. A very small ball is released from the hot air balloon at the instant when it is 1120 feet above the ground. Use a(t)=−32 ft/sec^2 as the acceleration due to gravity.
Required:
a. How many seconds after its release will the ball strike the ground?
b. At what velocity will it hit the ground?
Answer:
Here we only need to analyze the vertical motion of the ball.
First, because the ball is in the air, the only force acting on the ball will be the gravitational force (we are ignoring air resistance), then the acceleration of the ball is equal to the gravitational acceleration, so we have:
a(t) = -32 ft/s^2
where the negative sign is because this acceleration is downwards.
To get the velocity equation, we need to integrate over the time, so we get:
v(t) = (-32 ft/s^2)*t + V0
where V0 is the initial velocity of the ball. In this case, the initial velocity of the ball will be the velocity that the ball had when it was dropped, which should be the same as the velocity of the hot air ballon, so we have:
V0 = 12 ft/s
Then the velocity equation of the ball is:
v(t) = (-32 ft/s^2)*t + 12 ft/s
To get the position equation we integrate again:
p(t) = (1/2)*(-32 ft/s^2)*t^2 + (12 ft/s)*t + H0
Where H0 is the initial height. We know that the ball was released at the height of 1120 ft, then we have:
H0 = 1120 ft.
Then the position equation is:
p(t) = (-16 ft/s^2)*t^2 + (12 ft/s)*t + 1120ft
a) How many seconds after its release will the ball strike the ground?
The ball will strike the ground when its position equation is equal to zero, then we need to solve:
p(t) = 0 ft = (-16 ft/s^2)*t^2 + (12 ft/s)*t + 1120ft
This is just a quadratic equation, the solutions are given by Bhaskara's formula, so the solutions for t are:
[tex]t = \frac{- 12ft/s \pm \sqrt{(12 ft/s)^2 - 4*(-16ft/s^2)*(1120 ft)} }{2*(-16ft/s^2)}[/tex]
We can simplify that to get:
[tex]t = \frac{-12ft/s \pm 268ft/s}{-32ft/s^2}[/tex]
So we have two solutions:
[tex]t = \frac{-12ft/s + 268ft/s}{-32ft/s^2} = -8s[/tex]
[tex]t = \frac{-12ft/s - 268ft/s}{-32ft/s^2} = 8.75s[/tex]
The negative solution does not make sense, then the correct solution is the positive one.
We can conclude that the ball will hit the ground after 8.75 seconds.
b) At what velocity will it hit the ground?
We already know that the ball strikes the ground 8.75 seconds after it is released.
The velocity at it hits the ground is given by the velocity equation evaluated in that time:
v(8.75 s) = (-32 ft/s^2)*8.75s + 12 ft/s = -268 ft/s
Find the mean of the data in the pictograph below.
Answer:
12 sundaes
Step-by-step explanation:
Please help me, by completing this proof!
Answer:
Step-by-step explanation:
Statement Reasons
1). Line PQ is an angle bisector of ∠MPN D). Given
2). ∠MPQ ≅ ∠NPQ A). Definition of angle bisector
3). m∠MPQ = m∠NPQ F). Definition of congruent
angles.
4). m∠MPQ + m∠NPQ = m∠MPN C). Angle addition postulate
5). m∠MPQ + m∠MPQ = m∠MPN G). Substitution property of
equality
6). 2(m∠MPQ) = m∠MPN B). Distributive property
7). m∠MPQ = [tex]\frac{1}{2}(m\angle MPN)[/tex] E). Division property of equality
