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4 years ago

A dilation is a transformation whose preimage and image are...

Mathematics

1 answer:

grigory [225]4 years ago

4 0

The images will be similar, as dilation is merely increasing or decreasing the size of an image.

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IM BEGGING PLZ! I’ll mark brainliest! Plz!

Free_Kalibri [48]

Answer:

Step-by-step explanation:

SOHCAHTOA

Sine is opposite over hypotenuse

Cosine is adjacent over hypotenuse

Tangent is opposite over adjacent

In this context, opposite means the side opposite of the angle. Adjacent means the side that is touching the angle. And the hypotenuse is the longest side, which is usually visually obvious.

3.) tan(25) = x/31

Use a calculator for tan(25), then multiply both sides by 31 to get your x value.

x = 14.5

4.) cos(48) = 17/x

Use a calc to do 17/cos(48) to get the x value

x = 25.4

5.) sin(theta) = 17/18

Divide 17/18, then take the inverse sine of both sides with a calculator to get the theta value

theta = 70.8 degrees

6.) tan(theta) = 31/42

Divide 31/42, then take the inverse tangent of both sides with a calc.

theta = 36.4 degrees

7.) Your best bet is to draw a picture. For my case, Im going to imagine it. The angle is 35 degrees and the hypotenuse is 300 feet. The question is asking for the opposite of the angle. This calls for sine.

sin(35) = x/300

x = 172.1 feet

8 0

3 years ago

Please help me! No links as answers please. Find the expected value: Tim wins $3 if a coin toss shows heads or $2 if it shows ta

amm1812

The answer and the expected value is $2.50

4 0

3 years ago

One root of the equation x^2+6x+q=0 is twice the other root. Find q and the roots.

ella [17]

Answer:

q = 8

x₁ = -2, x₂ = -4

Step-by-step explanation:

For a quadratic equation ax² + bx + c, the sum of the roots is -b/a, and the product of the roots is c/a.

If the roots are x₁ and x₂, then:

-6/1 = x₁ + x₂

q/1 = x₁ x₂

Since we know one root is double the other, we can say x₂ = 2x₁. Plugging into the first equation and solving:

-6 = x₁ + 2x₁

-6 = 3x₁

x₁ = -2

Which means x₂ = -4. So the value of q must be:

q/1 = (-2) (-4)

q = 8

4 0

3 years ago

half of a pizza with broccoli and half with mushroom George 8 one-third of the broccoli part and one fourth of the mushroom part

HACTEHA [7]

He ate 7/12 of the pizza because 1x4=4, 1x3=3, 4+3=7, and 4x3=12.

Hope that helps.

4 0

3 years ago

An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as

Paul [167]

Answer:

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is $\frac{a}{\sqrt{3}}=R$ .

Upon substituting our given values, we will get:

$\frac{5x}{\sqrt{3}}=6r$

Let us solve for r.

$r=\frac{5x}{6\sqrt{3}}$

$\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}$

We know that area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4}s^2$ , where s represents side length of triangle.

$\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2$

The area within circle and outside the triangle would be difference of area of circle and triangle as:

$A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}$

We can make a common denominator as:

$A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}$

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Therefore, our required expression would be $A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$ .

7 0

3 years ago