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Vera_Pavlovna [14]

3 years ago

13

mann middle school has 1,000 students, 40 teachers, and 5 administrators. if the school grows to 1,200 students and the ratios a

re maintained, find the number of administrators that will be needed.

1 answer:

gregori [183]3 years ago

7 0

6 administrators are needed

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The function h(x)= x^2+3 and g(x)= x^2-6. If g(x) = h(x) +k, what is the value of k

trapecia [35]

Answer:

k = - 9

Step-by-step explanation:

h(x) = x² + 3

g(x) = x² - 6

g(x) = h(x) + k

x² - 6 = x² + 3 + k

-6 - 3 = k

k = - 9

8 0

2 years ago

Read 2 more answers

You work for a manufacturing company and have just completed the budget process for the upcoming business year. At the end of th

fiasKO [112]

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Step-by-step explanation:

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3 0

2 years ago

Evaluate 7 r − 15 s 7r− s 15 7, r, minus, start fraction, 15, divided by, s, end fraction when r = 3 r=3r, equals, 3 and s = 5

NeTakaya

Answer: 18

Step-by-step explanation:

Evaluate 7 r − 15 /s

Given that r = 3 and s = 5

7r - 15/s

Substitute 3 for r and 5 for s in the above expression

7(3) - 15/5

= 21 - 3

= 18

3 0

2 years ago

3=b+3<br><br> A. 0<br> B. 1<br> C. 9<br> D. 6

jok3333 [9.3K]

Your Answer Is...

A) 0.

Why?

This is correct because if 3 = b + 3, then ? + 3 = 3. 0 + 3 = 3, making the answer A) 0.

≈≈≈≈ Glad I Could Help, And Good Luck! ≈≈≈≈

My Name: AnonymousGiantsFan

7 0

2 years ago

Help!! Answer !!! about to run out of time in test!!

Gemiola [76]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the solution ~

Let's find the measure of hypotenuse first, by using Pythagoras theorem ;

$\qquad \sf \dashrightarrow \: h {}^{2} = {8}^{2} + {6}^{2}$

$\qquad \sf \dashrightarrow \: h {}^{2} = {36}^{} + {64}^{}$

$\qquad \sf \dashrightarrow \: h {}^{2} = 100$

$\qquad \sf \dashrightarrow \: h {}^{} = \sqrt{100}$

$\qquad \sf \dashrightarrow \: h {}^{} = {10}$

Now, let's find the asked values ~

$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{opposite \: side}{hypotenuse}$

$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{6}{10}$

$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{3}{5} \: or \: 0.6 \: units$

For Cos y :

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{adjcant \: side}{hypotenuse}$

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{6}{10}$

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{3}{5} \: or \: 0.6 \: units$

As we can see that both sin x and Cos y have equal values, therefore The required relationships is equality.

I.e Sin x = Cos y

Hope it helps ~

3 0

2 years ago