find all the missing measurement​

Answer:

This the right order:

4^2+2^2 = 20

5^2= 25

6^2-6 = 30

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

The focus here is the use of "Compounding interest rate" and these entails addition of interest to the principal sum of the deposit.

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

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Answer

the simplied form of the given question is 1

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  

Answer:

2x-30 + x + 40 = 180

3x + 10 = 180

3x = 170

x = 56 [tex]\frac{2}{3}[/tex]

Step-by-step explanation:

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.

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

Calculus

Differentiation

Integration

Integration 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

Reduction 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).

Step 3: Integrate Pt. 2

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

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

Answer:

litrally I don't understand what you are telling

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.

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

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

Considering the given function, we have that:

a) 28.31 years.

b) 0.3382 years a decade.

c) 2.6184.

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

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

Answer:

12 sundaes

Step-by-step explanation:

Answer:

its the same above

Step-by-step explanation:

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

Answer:

a+b=c+d/2

i cant understand what answer you want

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]

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.

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

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.

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).

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.

Answer:

The equation of the parabola with a focus at (2,4) and a directrix of y=8 is,

48-8y=(x-2)²

Answer:

Poggers

Step-by-step explanation:

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.