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3 years ago

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in a gym class the ratio of girls to total students is 4:7 there are 9 boys in the gym class how many students are in the gym cl

ass

1 answer:

nikklg [1K]3 years ago

3 0

4:7 = x/9+x. 36 + 4x = 7x. 36=3x. x=12. x=girls in class. x+9=all students in class.

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Plz help meee answer fast

bazaltina [42]

Answer: 85

Step-by-step explanation:

sorry if this is wrong did it quick. 20+30+35=85v :)

5 0

3 years ago

20 1/3 + x = -9 round your answer to the nearest hundredth

nadya68 [22]

Answer:

I would think it's X= -29 1/3

Step-by-step explanation:

You have 20 1/3 already, and to get to a negative when you have a positive number, to the positive you would need to add a larger negative. (If that makes any sense, it's hard to explain.) So you have to get rid of the 20 1/3 and you have to get to -9, so I would add a 9 to 20 1/3 to get 29 1/3, and make that number negative because then it will cancel the positive out. Making it -9.

3 0

3 years ago

At the movie theater,the Clayton family bought 4 fountain drinks and 3 bags of popcorn and paid $22.75.how much is a fountain dr

Anna35 [415]

<h3><u>Question:</u></h3>

At the movie theater , the Clayton family bought 4 fountain drinks and 3 bags of popcorn and paid $22.50. The Hanks family bought 5 fountain drinks and 2 bags of popcorn and paid $23.75. How much is a fountain drink? How much is a bag of popcorn ?

<h3><u>Answer:</u></h3>

Cost of 1 fountain drink is $ 3.75 and cost of 1 bag of popcorn is $ 2.5

<h3><u>Solution:</u></h3>

Let "f" be the price of 1 fountain drink

Let "p" be the price of 1 bag of popcorn

Given that Clayton family bought 4 fountain drinks and 3 bags of popcorn and paid $22.50

4 x price of 1 fountain drink + 3 x price of 1 bag of popcorn = 22.50

$4 \times f + 3 \times p = 22.50$

4f + 3p = 22.5 -------- eqn 1

Also given that Hanks family bought 5 fountain drinks and 2 bags of popcorn and paid $23.75

5 x price of 1 fountain drink + 2 x price of 1 bag of popcorn = 23.75

$5 \times f + 2 \times p = 23.75$

5f + 2p = 23.75 -----------------eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

2(4f + 3p = 22.5)

8f + 6p = 45 -------- eqn 3

Multiply eqn 2 by 3

3(5f + 2p = 23.75)

15f + 6p = 71.25 --------- eqn 4

<em><u>Subtract eqn 3 from eqn 4</u></em>

15f + 6p = 71.25

8f + 6p = 45

( - ) -------------------

7f = 26.25

f = 3.75

<em><u>Substitute f = 3.75 in eqn 1</u></em>

4(3.75) + 3p = 22.5

15 + 3p = 22.5

3p = 22.5 - 15

3p = 7.5

p = 2.5

Thus cost of 1 fountain drink is $ 3.75 and cost of 1 bag of popcorn is $ 2.5

8 0

3 years ago

100 points please help

Sav [38]

Answer:

${ \tt{(3x - 14) \degree = (2x + 10) \degree}} \\ { \rm{ \{alternate \: angles \}}} \\ \\ { \tt{3x - 2x = 10 + 14}} \\ \\ { \boxed{ \tt{ \: x = 24 \: }}}$

5 0

3 years ago

Read 2 more answers

If sinx = p and cosx = 4, work out the following forms :<br><br><br>

Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$

8 0

3 years ago