The two solids are similar, and the ratio between the lengths of their edges is
2:9. What is the ratio of their surface areas?
36 2
A. 64:729
B. 4:81
C. 4:18
D. 8:36

Step-by-step explanation:

4x−5y=6

−8x+10y=−12

Isolate x for 4x –5y = 6: x=6-5y/4

Substitute x= 6+5y/4

[-8 (6-5y/4) +10y=-12]

Simplify

[-12= -12 ]

The solutions to the system of equations are:

x=6+5y/4

Answer:

3/4_2/3 find the LCM= 1/12

Answer:

r = 50 degrees.

Step-by-step explanation:

r = 50 degrees ( vertical angles are congruent).

Answer:

With Graph and Without Graph.

Step-by-step explanation:

Without graphing calculator, you plug in your x-values into the equation

y = 16 -x^2 to solve for y-values.

f(x) = 16 - x^2

f(-4) = 16 - (-4)^2 = 16 - 16 =0

...

f(4) = 16 - (4)^2 = 16 - 16 =0

With Graphing Calculator.

The base of a solid is the region in the first quadrant bounded by the y-axis, the x-axis, the graph of y=ex, and the vertical line x=1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid?

Step-by-step explanation:

Answer: Sometimes I dont want to be happy.

[tex]~~~~1+ \cos \theta = 2 \cos^2 \theta \\\\\implies 2\cos^2 \theta -\cos \theta -1 = 0\\\\\implies 2 \cos^2 \theta -2\cos \theta + \cos \theta -1 = 0\\\\\implies 2 \cos \theta( \cos \theta -1) +(\cos \theta -1)=0\\\\\implies (\cos \theta -1)(2 \cos \theta +1)=0\\\\\implies \cos \theta = 1, ~~\cos \theta = -\dfrac 12\\\\\implies \theta = 2n\pi,~~~ \theta = 2n\pi \pm \dfrac{2\pi}3[/tex]

Answer:

y=6^x+0.612 - 4

Step-by-step explanation:

x = -4 ---> horizontal asymptote

m = 6 ---> use points (0, -1) and (1, 5)

Parent function of the graph: [tex]y=b^x[/tex]

Our equation: [tex]y=6^x[/tex]

Add alterations:

Final equation: y=6^x+0.612 - 4

Answer:

Use   SOH  CAH TOA  to rember how the trig function fit on the triangle

Step-by-step explanation:

we are given  the Hypotenuse and the Opposite  or   H and O   look for the trig function with each of those,  SOH is good

Sin Ф =  Opp /  Hyp,    or   SOH

now plug in what you are given

Sin Ф = 5 / 8

use inverse trig fuction to find the angle

arcSin ( Sin Ф) = arcSin ( 5/8)

trig functions cancel out

Ф = arcSin(5/8)

I'm using my calculator to find the arcSin(5/8)

Ф=38.6821...°

also make sure you know if your calculator is in degrees or radians.  

Ф=38.68°   to the nearest hundredth  :)

Answer:

∠A = 38.68°

Step-by-step explanation:

The side opposing ∠A and the hypotenuse are given.

Therefore, take the inverse sin function of ∠A.

Use logarithms to solve for a.

[tex]3^a = 2 \implies \log_3(3^a) = a\log_3(3) = \log_3(2) \implies a = \log_3(2)[/tex]

Similarly, for b and x.

[tex]3^b = 5 \implies b = \log_3(5)[/tex]

[tex]3^x = 150 \implies x = \log_3(150)[/tex]

Factorize 150:

150 = 2 • 3 • 5²

Then we can expand log₃(150) using the product-to-sum and exponent property,

[tex]\log_3(150) = \log_3(2\times3\times5^2) = \log_3(2) + \log_3(3) + \log_3(5^2)[/tex]

[tex]\implies \log_3(15) = \log_3(2) + 1 + 2 \log_3(5) \iff \boxed{x = 1 + a + 2b}[/tex]

Answer:

$18.60

Step-by-step explanation:

The total amount spent is $372, and this can be divided by 20. Each costume would cost $18.60.

Answer:

The length of the arc is 8*pi cm. An arc that subtends a central angle of 120 degrees has a length of 120/360 = 1/3 the length of the total circumference of the circle. We know the entire circumference of the circle is 2πr, which in this case is 2π*12 = 24π.

Step-by-step explanation:

sorry I only have the 8

Answer:

$3096

Step-by-step explanation:

There are 12 months in 1 year.

3 × 12 = 36

There are 36 months in 3 years.

$86 × 36 = $3096

Answer:

3096

Step-by-step explanation:

I just completed the quiz.

Answer:

See below ~

Step-by-step explanation:

Given

Solving

Take Equation 1 and rewrite it so that it is equal to x.

Take Equation 2 and rewrite it so that it is equal to z.

Now, substitute these values in Equation 3.

Substitute the value of y in Equation 1.

Substitute the value of y in Equation 2.

Add the values of x, y, and z together.

Answer:

  30 mL

Step-by-step explanation:

You are being asked to solve the given equation for the value of x.

__

  0.10x +0.15(50 -x) = 0.12(50) . . . . given

  -0.05x = -0.03(50) . . . . . subtract 0.15(50), combine terms

  x = 30 . . . . . . . . . . . divide by -0.05

Bruce should use 30 mL of the 10% alcohol solution.

Answer:

ir

Step-by-step explanation:

irregular

regular

what do you notice different?

the both have r, e, g, u, l, a, r, but one has ir at the beginning.

Answer:

x = -2 and 6

Step-by-step explanation:

To find vertical asymptote, set the denominator equal to 0 and solve for x. See the guidelines below for determining VA

[tex]y=\frac{x-5}{x^{2} -4x-12}[/tex]

[tex]x^{2} -4x-12=0[/tex]

[tex]x^{2} -6x+2x-12=0[/tex]

[tex]x(x-6)+2(x-6)=0[/tex]

[tex](x+2)(x-6)=0[/tex]

[tex]x=-2,6[/tex]

Hope this helps and God bless!

The linear function that gives the candle's height, y, after x hours, is defined by:

y = -0.25x + 11.

A linear function is modeled by:

y = mx + b

In which:

In this problem, in 12 hours, the height decayed 3 inches, hence the slope is given by:

m = -3/12 = -0.25.

So:

y = -0.25x + b.

After 4 hours, the height was of 10 inches, hence we use this to find b:

10 = -0.25(4) + b

b = 11

Thus the equation is given by:

y = -0.25x + 11.

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Answer:

1. Triangle: B = 47.0° , C = 103.05° , c = 2.53 cm

2. Lake: c = 1105.31 ft

Step-by-step explanation:

Law of Sine Formula:[tex]\frac{sin(A)}{A} = \frac{sin(B)}{B} = \frac{sin(C)}{C}[/tex]

Given: A = 30° , a = 1.3 cm , b = 1.9 cm

[tex]\frac{sin(30)}{1.3} = \frac{sin(B)}{1.9} = \frac{sin(C)}{C}[/tex]

Solving for sin(B). Cross Multiply.

[tex]\frac{sin(30)}{1.3} = \frac{sin(B)}{1.9}\\1.3*sin(B)=1.9*sin(30)\\sin(B)=\frac{1.9*sin(30)}{1.3} \\[/tex]

B = sin^-1( [tex]\frac{1.9*sin(30)}{1.3}[/tex] )

B ≈ 46.9509202

B = 47.0°

Solve for C°

A° + B° + C° = 180°

30° + 46.95° + C° = 180°

C° = 180° - 30° - 46.95°

C° = 103.05°

Solve for sin(C)

[tex]\frac{sin(30)}{1.3} = \frac{sin(103.05)}{C}\\[/tex]

Cross Multiply

[tex]C*sin(30)=1.3*sin(103.05)\\C=\frac{1.3*sin(103.05)}{sin(30)}[/tex]

C ≈ 2.532850806

C = 2.53 cm

Law of Cosine Formula: [tex]c^2=a^2+b^2-2*a*b*cos(C)[/tex]

Given: a = 850 ft , b = 960 ft ,  C=75°

Solve for c.

[tex]c^2=a^2+b^2-2*a*b*cos(C)\\c^2=(850ft)^2+(960ft)^2-2*(850ft)*(960ft)*cos(75)\\\\c=\sqrt{(850ft)^2+(960ft)^2-2*(850ft)*(960ft)*cos(75)\\} \\[/tex]

c ≈ 1105.308698

c = 1105.31 ft

Answer:

$77

Step-by-step explanation:

The answer is $77 because first you need to find how much money was discounted by doing 70-49 to get 21. Then you need to find how much percent 21 is of 70 by doing 21/70, then you would get 0.3 which is 30% since you have to multiply it by 100. This means that there is a 30% discount. Then you would do 0.3*110=33. This means that the 30% discount takes away $33. So 110-33=77. The answer is $77.

Answer:

A. r = 6 sin theta

Step-by-step explanation:

Given equation is: [tex]x^2+y^2-6y=0[/tex]....(1)

Using the formulae that link Cartesian to Polar coordinates.

[tex]x=r\cos\theta \: and \: y = r\sin\theta[/tex]

Plugging the values of x and y in equation (1), we find:

[tex](r\cos\theta)^2+(r\sin\theta)^2-6(r\sin\theta)=0[/tex]

[tex]\implies r^2\cos^2\theta+r^2\sin^2\theta-6r\sin\theta=0[/tex]

[tex]\implies r^2(\cos^2\theta+\sin^2\theta)=6r\sin\theta[/tex]

[tex]\implies r^2(1)=6r\sin\theta[/tex]

[tex](\because \cos^2\theta+\sin^2\theta=1)[/tex]

[tex]\implies\frac{ r^2}{r}=6\sin\theta[/tex]

[tex]\implies\huge{\purple{ {r}=6\sin\theta}}[/tex]

Answernone

none

none

Step-by-step explanation:

Answer: rectangular prism, triangular prism, 6

Step-by-step explanation:

The area of base of the triangular prism can be calculated by the half of the product of the height of the triangular base of prism and base of prism.

The area of base of the rectangular prism can be calculated by the product of the length of the rectangular base of prism and the breadth of prism.

The area of the bottom section of the second floor = 2( area of the front part of the wall + area of the side of the wall)

= 2( 20*h + 25*h)

=90h

The front and back of top section of second floor = 2* area of the front wall=2*(1/2*20*(h+4))=20h+80

Total area of the second floor=total area of the window+total area to be painted

⇒Total area of the second floor= 728 + (2* area of the window)

⇒Total area of the second floor= 728 + (2*3*2)

⇒area of front and back of top section of second floor + The area of the bottom section of second floor = 728+12

⇒area of front and back of top section of second floor + The area of the bottom section of second floor = 740

⇒(20h+80)+90h=740

⇒110h+80=740

⇒110h=740-80

⇒110h=660

⇒h=660/110

⇒h=6 feet

Therefore the value of h is 6 feet.

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By definition of cosine and sine,

cos(θ) = x/r

sin(θ) = y/r

so that

cos²(θ) + sin²(θ) = (x/r)² + (y/r)²

… = x²/r² + y²/r²

… = (x² + y²)/r²

… = r²/r²

… = 1

To that end, I would say

• [blank1] = "Division property of equality"

That is, we divide both sides of an equation by the same number and equality still holds since r ≠ 0

• [blank2] = "Definition of cos"

• [blank3] = "cos²(θ) = x²/r²"

• [blank4] = "Defintion of sin"

• [blank5] = "sin²(θ) = y²/r²"

• [blank6] = "Simplify"

More specifically, x² + y² = r² is given, so

x²/r² + y²/r² = (x² + y²)/r² = r²/r² = 1

Substitute [tex]x\mapsto\sqrt{1-x^2}[/tex], which transforms the integral to

[tex]\displaystyle \int_0^1 \ln^{2k} \left(\frac{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}\right) \, dx = \int_0^1 \ln^{2k}\left(\frac{\ln\left(\frac{1-x}{\sqrt{1-x^2}}\right)}{\ln\left(\frac{1+x}{\sqrt{1-x^2}}\right)}\right) \frac{x}{\sqrt{1-x^2}} \, dx[/tex]

and factoring [tex]\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}[/tex] reduces this to

[tex]\displaystyle = \int_0^1 \ln^{2k}\left(\frac{\ln\left(\sqrt{\frac{1-x}{1+x}}\right)}{\ln\left(\sqrt{\frac{1+x}{1-x}}\right)}\right) \frac x{\sqrt{1-x^2}} \, dx[/tex]

The inner logarithms differ only by a sign, so that

[tex]\displaystyle = \int_0^1 \ln^{2k}(-1) \frac x{\sqrt{1-x^2}} \, dx[/tex]

Using the principal branch of the complex logarithm, we have

[tex]\ln(-1) = \ln|-1| + i\arg(-1) = i\pi[/tex]

and hence

[tex]\displaystyle \int_0^1 \ln^{2k} \left(\frac{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}\right) \, dx = (i\pi)^{2k} \underbrace{\int_0^1 \frac x{\sqrt{1-x^2}} \, dx}_{=1} = \boxed{(-\pi^2)^k}[/tex]

where I assume k is an integer.

Answer:

umm 9?

Step-by-step explanation: