Find the amplitude and period of the function y=3/2 cos(4/3x)

The length of a leg of an isosceles right triangle is 20 inches. Find the length of the hypotenuse in radical form

Over [174]

Answer: H = 20 $\sqrt{2}$

Step-by-step explanation:

To find the hypotenuse , we will use Pythagoras theorem, which states that

The square of the hypotenuse = the square of the adjacent + the square of the opposite.

Since , we are dealing with isosceles right triangle , this means that :

both the adjacent and the opposite are equal, using the formula to find the hypotenuse , let H represent the hypotenuse , then

$H^{2}$ = $20^{2}$ + $20^{2}$

$H^{2}$ = 400 + 400

$H^{2}$ = 800

Take the square root of both sides

H = $\sqrt{800}$

$\sqrt{800}$ can be written as $\sqrt{400}$ x $\sqrt{2}$

Therefore :

H = $\sqrt{400}$ x $\sqrt{2}$

H = 20x $\sqrt{2}$

H = 20 $\sqrt{2}$

6 0

3 years ago

There are two spinners. Each spinner has 10 equal sectors labeled with the numbers 1 through 10.

murzikaleks [220]

Answer:

There are 4 prime numbers from 1 to 10: 2, 3, 5, and 7. This means that there are 4 ways to choose the first outcome. There are 5 composite numbers from 1 to 10: 4, 6, 8, 9, and 10 (1 is just 1). So, you just multiply the two numbers together because each depends on each: 4 * 5 = 20 outcomes.

Step-by-step explanation:

please give me a brainliest answer

8 0

3 years ago

Simplify so that there are only positive exponents (3^9/3^-5)

fomenos

Answer:

3^14

Step-by-step explanation:

(3^9/3^-5)

We know that when we have a^b / a^c = a^(b-c)

3^(9 - -5)

3^(9+5)

3^14

5 0

3 years ago

2 out of every 7 fans at the Wombats vs. Wallabies game we're season ticket holders. If 2,800 fans attended the game, how many d

White raven [17]

It would be 800
2/7 = 2800/x

7 0

3 years ago

Someone please help me!!!!

BARSIC [14]

Answer: see proof below

Step-by-step explanation:

Use the Double Angle Identity: cos 2A = 2cos²A - 1

Use the Difference Identity: cos (A - B) = cosA · cosB + sinA · sinB

Use the Unit Circle to evaluate: cos (π/2) = 0 & sin (π/2) = 1

Proof LHS → RHS

$\text{Given:}\qquad \qquad \qquad \qquad 2\cos^2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)-1\\\\\\\text{Double Angle Identity:}\quad \cos2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)\\\\\\\text{Simplify:}\qquad \qquad \qquad \quad \cos\bigg(\dfrac{\pi}{2}-A\bigg)\\\\\\\text{Difference Identity:}\qquad \cos\dfrac{\pi}{2}\cdot \cos A+\sin \dfrac{\pi}{2}\cdot \sin A\\\\\\\text{Unit Circle:}\qquad \qquad \qquad 0\cdot \cos A+1\cdot \sin A\\\\\\\text{Simplify:}\qquad \qquad \qquad \qquad \sin A$

LHS = RHS: sin A = sin A $\checkmark$

4 0

4 years ago

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