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3 years ago

8

It should take 5.22 minutes to clean a shower. You have 5 showers that need to be cleaned. How many minutes should it take for y

ou to clean all 5?

2 answers:

PSYCHO15rus [73]3 years ago

8 0

All you have to do is plug 5.22*5 into your calculator and you should get 26.1.

Mariulka [41]3 years ago

8 0

26.1 minutes to clean

You might be interested in

Help please!! Quick

Finger [1]

Step-by-step explanation:

Solve for a:

Step 1: Multiply both sides by b.

b=a

Step 2: Flip the equation.

a=b

---------------------------

Solve for b:

Step 1: Multiply both sides by b.

b=a

4 0

3 years ago

The height of a square pyramid-shaped gift box is one half the length of

Schach [20]

Answer:

15 inches

Step-by-step explanation:

Volume of a squre based pyramid :

V = a²(h/3)

a = base edge ; h = height

h =1/2 a

V = 4500

4500 = a² * (1/2a) ÷ 3

4500 = a² * a/2 * 1/3

4500 = a³/2 * 1/3

4500 = a³ / 6

27000 = a³

27000 = a³

Take the cube root of both sides

(27000)⅓ = a

30 = a

Recall:

h = 1/2a

h = 1/2(30)

h = 15 inches

5 0

2 years ago

URGENT!!!<br> Which function is shown in the graph?

katrin2010 [14]

Answer:

<h2>it's b</h2><h2>you can go with it</h2>

...........

5 0

2 years ago

Given g(x) = x4 + 2x3 - 7x2 - 8x + 12. When g(x) is divided by x - 1, which conclusion about g(x) is true?

PSYCHO15rus [73]

Answer:

g(1) = 0

There is a remainder of 10 so you can automatically cross out 3 and 4.

7 0

2 years ago

Read 2 more answers

Please help me to prove this!

Ymorist [56]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π → A + B = π - C

→ B + C = π - A

→ C + A = π - B

→ C = π - (B + C)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → Middle:</u>

$\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)$

$\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)$

$\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)$

$\text{Sum/Difference:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)$

$\text{Double Angle:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)$

$\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]$

$\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]$

$\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)$

$\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)$

LHS = Middle $\checkmark$

<u>Proof Middle → RHS:</u>

$\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)$

Middle = RHS $\checkmark$

3 0

3 years ago