Which series represents this situation? 1+1*7+1*7^ 2 +...1*7^ 6; 1+1*7+1*7^ 2 +...1*7^ 7; 7+1*7+1*7^ 2 +... 1*7^ 6; 7+1*7+1*7^ 2 +...1*7^ 7

Answer:

[tex]\Large \boxed{x_1=\frac{9+\sqrt{51} }{3} \ \ ; \ \ x_2=\frac{9-\sqrt{51} }{3} }[/tex]

Step-by-step explanation:

[tex]\displaystyle \Large \boldsymbol{} 3x(x+6)=-10 \\\\3x^2+18x=-10 \\\\3x^2+18x+10=0 \\\\D=324-120=204 \\\\ x_{1;2}=\frac{18\pm2\sqrt{51} }{6} =\frac{9\pm\sqrt{51} }{3}[/tex]

Solution :

Case I :

If Collen is late on [tex]0[/tex] out of [tex]5[/tex] days.

[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} $[/tex]

[tex]$=\frac{1}{32}[/tex]

Case II :

When Collen is late on [tex]1[/tex] out of [tex]5[/tex] days.

[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times ^5C_1$[/tex]

[tex]$=\frac{1}{32} \times 5$[/tex]

[tex]$=\frac{5}{32}[/tex]

Case III :

When Collen was late on [tex]2[/tex] out of [tex]5[/tex] days.

[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times ^5C_2$[/tex]

[tex]$=\frac{1}{32} \times 10$[/tex]

[tex]$=\frac{5}{16}[/tex]

Therefore, the [tex]\text{probability}[/tex] that Collen will arrive late to work no more than [tex]\text{twice}[/tex] during a [tex]\text{five day workweek}[/tex] is :

[tex]$=\frac{1}{32} + \frac{5}{32} + \frac{5}{16} $[/tex]

[tex]$=\frac{1}{2}$[/tex]

Answer:

g/c =m

Step-by-step explanation:

g=4cm-3cm

Combine like terms

g = 1cm

g =cm

Divide by c

g/c = cm/c

g/c =m

Answer:

C. 7 units

Step-by-step explanation:

The given parameters are;

The length of the chord of the circle, [tex]\overline{AC}[/tex] = 14 units

The orientation of the radius and the chord = The radius is perpendicular to the chord

We have in ΔAOC, [tex]\overline{AO}[/tex] = [tex]\overline{OC}[/tex] = The radius of the circle

[tex]\overline{OB}[/tex] ≅ [tex]\overline{OB}[/tex]  by reflexive property

The angle at point B = 90° by angle formed by the radius which is perpendiclar to the chord [tex]\overline{AC}[/tex]

ΔAOB and ΔCOB are right triangles (triangles having one 90° angle)

[tex]\overline{AO}[/tex] and [tex]\overline{OC}[/tex] are hypotenuse sides of ΔAOB and ΔCOB respectively and [tex]\overline{OB}[/tex] is a leg to ΔAOB and ΔCOB

Therefore;

ΔAOB ≅ ΔCOB, by Hypotenuse Leg rule of congruency

Therefore;

[tex]\overline{AB}[/tex] ≅ [tex]\overline{BC}[/tex] by Congruent Parts of Congruent Triangles are Congruent, CPCTC

[tex]\overline{AB}[/tex] = [tex]\overline{BC}[/tex] by definition of congruency

[tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] by segment addition postulate

∴ [tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] =  [tex]\overline{AB}[/tex] + [tex]\overline{AB}[/tex] = 2 ×  [tex]\overline{AB}[/tex]

∴  [tex]\overline{AB}[/tex] = [tex]\overline{AC}[/tex]/2

[tex]\overline{AB}[/tex] = 14/2 = 7

[tex]\overline{AB}[/tex] = 7 units.

Answer:

7 units

Step-by-step explanation:

Answer:

The probability that you won't be a hobbit=

[tex] \frac{2}{3} [/tex]

The probability that you won't choose an umbrella=

[tex] \frac{4}{5} [/tex]

Step-by-step explanation:

The characters are 3, minus the hobbit gives you 2. So the probability of that=

[tex] \frac{2}{3} [/tex]

The weapons are 5, minus the umbrella is 4. So the probability of that=

[tex] \frac{4}{5} [/tex]

Answer:

224 ft

Step-by-step explanation:

The tree makes a 90° angle with the ground, so we have a right triangle.

For the 32° angle, the height of the tree is the opposite leg. The length of the shadow on the ground is the adjacent leg.

The trig ratio that relates the opposite leg to the adjacent leg is the tangent.

Let the length of the shadow of the tree on the ground be x.

tan A = opp/adj

tan 32° = 140 ft/x

x = 140 ft/tan 32°

x = 224 ft

Answer: 224 ft

Answer:

2020³-1/4041

Step-by-step explanation:

area of the parallelogram is: b x h

12 x 4 = 48 (d)

9514 1404 393

Answer:

  about 79 ft

Step-by-step explanation:

The relevant trig relation is ...

  Tan = Opposite/Adjacent

In this scenario, the angle is 23°, the adjacent side is the distance to the building, and the opposite side is the building height. Then we have ...

  height = tan(23°)·(186 ft) ≈ 78.95 ft ≈ 79 ft

Answer:

24 units ²

Step-by-step explanation:

In this problem, we are given the circumference of a triangle (after finding the perimeter) and want to find the area of a circle with that circumference. Since the area of a circle is a function based on its radius, we can use the circumference to find the radius to find the area.

First, we can figure out the perimeter of the triangle, which is equal to the sum of its sides. The perimeter is 6+4+7.21 = 17.21 units.

Next, the circumference of a circle is equal to π * diameter = π * 2 * radius. Using 3.14 for π and r for radius, we get

3.14 * 2 * r = 17.21

6.28 * r = 17.21

divide both sides by 6.28 to isolate r

r ≈ 2.74

Furthermore, to find the area from the radius, we can use

area = πr². Plugging 2.74 in for r, we get

2.74² * 3.14 = area

≈23.6, rounding up to 24 units ²

The circumference is equal to the perimeter of the triangle. Then the area of the circle is 23.58 or 24 square units. Option B is correct.

It is a locus of a point drawn equidistant from the center. The distance from the center to the circumference is called the radius of the circle.

A triangle has sides that measure 4 units, 6 units, and 7.21 units.

Then the perimeter of the triangle will be

Perimeter = 6 + 4 + 7.21

Perimeter = 17.21

The circumference is equal to the perimeter of the triangle. Then the radius will be

[tex]2\pi r = 17.21\\\\r = 2.74[/tex]

Then the area of the circle will be given as

[tex]\rm Area = \pi r^2 \\\\Area = \pi \times 2.74^2 \\\\Area = 23.58[/tex]

Thus, the area of the circle is 23.58 square units.

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Answer:

45 degrees

Step-by-step explanation:

the 2 angles are supplementary (add up to 180 degrees)

180-135=45

Answer:

12 minutes

Step-by-step explanation:

Given

[tex]s_1 = 10mi/h[/tex] --- walk speed

[tex]s_2 = 20mi/h[/tex] --- jog speed

[tex]d = 2\ miles[/tex] --- distance

Required

The time to walk the given distance

Time is calculated as:

[tex]Time = \frac{distance}{speed}[/tex]

In this case, the speed is the speed at which she walks.

So, we have:

[tex]Time = \frac{d}{s_1}[/tex]

Substitute known values

[tex]Time = \frac{2mi}{10mi/h}[/tex]

[tex]Time = 0.2hr[/tex]

Convert to minutes

[tex]Time = 0.2 * 60mins[/tex]

[tex]Time =12mins[/tex]

Answer:

2

y = 3x-15

y=5x-35

x1=5

x2=7

distance between = 7-5 = 2

Step-by-step explanation:

2x+54=8x

So. The Answer is 7

Answer:

67°

Step-by-step explanation:

let the base angle=x

x+x+46=180

2x=180-46=134

x=134/2=67

The Pascal triangle terms and binomial are related by the binomial theorem fomrula:

The row of the Pascal triangle is the same as

where n is the same as the row of the Pascal triangle.

For example,

.

is the same as the 2nd row of the Pascal triangle.

.

Also know the leading coefficient and last term degree will be the nth row of the Pascal triangle.

The first term, a will be 1 less than the previous terms and the second term, b will be 1 more than the previous terms. For example, look at

.

Another example is

.

The third row of Pascal Triangle

.

.

Pascal's triangle provides a systematic way to determine the coefficients of binomial expansions, making it a valuable tool in algebra and combinatorics.

Binomial expansions are directly related to Pascal's triangle.

Pascal's triangle is a triangular arrangement of numbers in which each number is the sum of the two numbers directly above it. The first and last numbers in each row of Pascal's triangle are always 1.

The coefficients of the terms in a binomial expansion correspond to the numbers in Pascal's triangle. Each row of Pascal's triangle represents the coefficients of the corresponding power of the binomial.

For example, in the expansion of (a + b)², the coefficients are 1, 2, and 1. These coefficients can be found in the third row of Pascal's triangle: 1, 2, 1. Similarly, in the expansion of (a + b)³, the coefficients are 1, 3, 3, and 1, which can be found in the fourth row of Pascal's triangle: 1, 3, 3, 1.

In summary, Pascal's triangle provides a systematic way to determine the coefficients of binomial expansions, making it a valuable tool in algebra and combinatorics.

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if you like my answer then please mark me as brainliest

Answer:

y''=(4-3t)/[16 t^3 (2-t)^(3/2)]

Step-by-step explanation:

x = t², y = (2 - t)^1/2

dy/dx=dy/dt×dt/dx by chain rule

dy/dt=1/2 (2-t)^(1/2-1) × (-1)

dy/dt=-1/2 (2-t)^(-1/2)

dy/dt=-1/[2(2-t)^(1/2) ]

dx/dt=2t

dy/dx=-1/[2(2-t)^(1/2) ] × 1/[2t]

dy/dx=-1/[4t (2-t)^(1/2) ]

We need to find the second derivative now.

That is we calculate d/dt(dy/dx in terms of t) then divide by derivative of x in terms of t).

dy/dx=-1/[4t (2-t)^(1/2) ]

Let's find derivative of this with respect to t.

d/dt(dy/dx)=

[0[4t (2-t)^(1/2)]-(-1)(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [4t (2-t)^(1/2) ]^2

Let's simplify

d/dt(dy/dx)=

[(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [4t (2-t)^(1/2) ]^2

Continuing to simplify

Apply the power in the denominator

d/dt(dy/dx)=

[(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [16t^2 (2-t) ]

Multiply by (2-t)^(1/2)/(2-t)^(1/2):

d/dt(dy/dx)=

[(4(2-t)+-4t(1/2)]/ [16t^2 (2-t)^(3/2)]

Distribute/multiply:

d/dt(dy/dx)=

[(8-4t+-2t)]/ [16t^2 (2-t)^(3/2)]

Combine like terms:

d/dt(dy/dx)=

[(8-6t)]/ [16t^2 (2-t)^(3/2)]

Reducing fraction by dividing top and bottom by 2:

d/dt(dy/dx)=

[(4-3t)]/ [8t^2 (2-t)^(3/2)]

Now finally the d^2 y/dx^2 in terms of t is

d/dt(dy/dx) ÷ dx/dt=

[(4-3t)]/ [8t^2 (2-t)^(3/2)] ÷ 2t

d/dt(dy/dx) ÷ dx/dt=

[(4-3t)]/ [8t^2 (2-t)^(3/2)] × 1/( 2t)

d/dt(dy/dx) ÷ dx/dt=

[(4-3t)]/ [16t^3 (2-t)^(3/2)]

Or!!!!!!

x = t², y = (2 - t)^1/2

Since t>0, then t=sqrt(x) or x^(1/2).

Make this substitution into the equation explicitly solved for y:

y = (2 - x^1/2)^1/2

Differentiate:

y' =(1/2) (2 - x^1/2)^(-1/2) × -1/2x^(-1/2)

y'=-1/4(2 - x^1/2)^(-1/2)x^(-1/2)

y'=-1/4(2x-x^3/2)^(-1/2)

Differentiate:

y''=1/8(2x-x^3/2)^(-3/2)×(2-3/2x^1/2)

y''=(2-3/2x^1/2)/[8 (2x-x^3/2)^(3/2)]

Replace x with t^2

y''=(2-3/2t)/[8 (2t^2-t^3)^(3/2)]

Multiply top and bottom by 2

y''=(4-3t)/[16 (2t^2-t^3)^(3/2)]

Factor out t^2 inside the 3/2 power factor:

y''=(4-3t)/[16 (t^2)^(3/2) (2-t)^(3/2)]

y''=(4-3t)/[16 t^3 (2-t)^(3/2)]

Answer:

15

Step-by-step explanation:

AB+BC=AC

2X+3+(4X-11)

6X-8=28

6x= 36

x=6

then ab= 2(6)+3

=15

bc= 4(6)-11

=13

ac=ab+bc

=15+13

=28

[tex] \frac{ \sin(20 \degree) }{ \cos(70\degree) } + \frac{ \cos(35\degree) }{ \sin( 65\degree)} [/tex]

[tex] \frac{ \sin(20 \degree) }{ \cos(70\degree) } + \frac{ \cos(35\degree) }{ \sin( 65\degree)} [/tex]

[tex] \sin( \theta) = \cos(90 - \theta) \\ \\ \cos( \theta) = \sin(90 - \theta) [/tex]

So putting down the value

[tex] \frac{ \cos(90 - 20 \degree) }{ \cos(70\degree) } + \frac{ \sin(90 - 35\degree) }{ \sin( 65\degree)} [/tex]

[tex] \frac{ \cos(70\degree) }{ \cos(70\degree) } + \frac{ \sin(65\degree) }{ \sin( 65\degree)} [/tex]

[tex]\frac{\cancel{\cos(70\degree)}}{ \cancel{\cos(70\degree)}} + \frac{\cancel{\sin(65\degree)}}{\cancel{\sin( 65\degree)}} [/tex]

[tex] \frac{1}{1} + \frac{1}{1} \\ 1 + 1 = 2 \: \: ans[/tex]

Answer:

Linear systems are systems of equations in which the variables are never multiplied with each other but only with constants and then summed up

Answer:

8

Step-by-step explanation:

x = centesimal angle (100 degrees in a right angle)

y = sexagesimal angle (90 degrees in a right angle)

x - y = 15

x×90/100 = y

x×9/10 = y

x - x×9/10 = 15

10x/10 - 9x/10 = 15

x/10 = 15

x = 150

y = 150×9/10 = 15×9 = 135

using centesimal system :

the sum of all external angles in a polygon is 400 degrees (a full circle).

one external angle is the complement of one internal angle to 200 degrees = 200-150=50 degrees.

to find the number of sides of the polygon we need to find the number of angles or corners. and that is how many external angles fit into the full circle.

n = 400/50 = 8

the polygon has 8 sides

Answer:

The width of the confidence interval based on the 784 women would be narrower than the  confidence interval based on the 327 women.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

From this, we have that the margin of error, and also the width, is inversely proportional to the sample size, that is, a larger sample size leads to a smaller margin of error and a narrower interval.

How would the width of the new confidence interval compare to the width of the  confidence interval based on the 327 women in the armed forces?

New interval: 784

Old interval: 327

Sample size increased, so the new interval will be narrower, and the correct answer is:

The width of the confidence interval based on the 784 women would be narrower than the  confidence interval based on the 327 women.

Answer:

Mike had 72 stamps

Ken had 120 stamps

Step-by-step explanation:

Given

[tex]M \to Mike[/tex]

[tex]K \to Ken[/tex]

[tex]\frac{1}{5} * K = \frac{1}{3} * M[/tex]

[tex]K + 24 = 3 * ( M - 24)[/tex]

Required

Find K and M

Make K the subject in: [tex]K + 24 = 3 * ( M - 24)[/tex]

[tex]K = 3 * ( M - 24) - 24[/tex]

Substitute [tex]K = 3 * ( M - 24) - 24[/tex] in [tex]\frac{1}{5} * K = \frac{1}{3} * M[/tex]

[tex]\frac{1}{5} * [3 * ( M - 24) - 24] = \frac{1}{3} * M[/tex]

Open brackets

[tex]\frac{1}{5} * [3M - 72 - 24] = \frac{1}{3} * M[/tex]

[tex]\frac{1}{5} * [3M -96] = \frac{1}{3} * M[/tex]

Multiply both sides by 15

[tex]3* [3M -96] = 5 * M[/tex]

[tex]9M -288 = 5M[/tex]

Collect like terms

[tex]9M -5M= 288[/tex]

[tex]4M= 288[/tex]

Divide both sides by 4

[tex]M= 72[/tex]

Substitute [tex]M= 72[/tex] in [tex]K = 3 * ( M - 24) - 24[/tex]

[tex]K = 3 * (72 - 24) - 24[/tex]

[tex]K = 3 * 48 - 24[/tex]

[tex]K = 120[/tex]

The total distance the particle travels from t=0 seconds to t=4 seconds would be 11.33 meters.

Used the concept of integration that states,

In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts.

Given that,

A particle moves along a line with a velocity v(t) = t² - t - 6, measured in meters per second.

Now the total distance the particle travels from t=0 seconds to t=4 seconds is,

D = ∫₀⁴ |(t² - t - 6)| dt

D =  ∫₀⁴ (t²) dt - ∫₀⁴ (t) dt -  ∫₀⁴ (6) dt

D = (t³/3)₀⁴ - (t²/2)₀⁴ - 6 (t)₀⁴

D =| (64/3) - (16/2) - 6 (4)|

D = | (64/3) - 8 - 24 |

D = | (64/3) - 32|

D = 11.33 meters

Therefore, the total distance is 11.33 meters.

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Answer:

Thamkhao

Step-by-step explanation:

Step-by-step explanation:

LCM of 7 and 5 is 35

now will make the denominator equal,

10+7/35

So, Ram's work = 10/35

Shyam's work = 7/35

so

Ram > Shyam

hope it helps

Answer:

4x-7=-5x²

or, 5x²+4x-7=0

so the standard form of the equation is,

5x²+4x-7=0

or, f(x) = 5x²+4x-7

Answered by GAUTHMATH