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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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<img src="https://tex.z-dn.net/?f=%20-%2030%20%3D%202%28x%20-%2097%29" id="TexFormula1" title=" - 30 = 2(x - 97)" alt=" - 30 = 2

Korolek [52]

Hey there!

In order to find out the answer to this question, we can distribute

$-30 = 2(x - 97) \\ \\ \\ 2(x) = 2x \\ \\ 2(97) = 197$

Flip your problem around!

$2x - 194 = -30$

Add by $194$ on each of your sides

Like: $2x - 194 + 194 \\ \\ \\ -30 + 194$

Cancel out: $-194 + 194$ because they would give you the outcome of $0$

Keep: $-30 + 194 [/tex] because that'll give you: [tex]164$

Your problem becomes: $2x = 164$

Now, we have to divide by $2$ on each of the sides

Like: $\frac{2x}{2} = \frac{164}{2}$

Cancel: $\frac{2x}{2}$ because it will give you a outcome of $1$

Keep: $\frac{164}{2}$ because this will give you your answer

$Answer: x = 82$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

7 0

3 years ago

Solve 3/5=8/y. Round to the nearest tenth. 15.2 9.8 13.3 18.8

Sidana [21]

3/5 = 3/y
40 = 3y
Divide both by 3
Answer= 13.3

4 0

4 years ago

During a period of 45 days, Delynn played soccer on 3/5 of the days. On how many days did she play soccer?

Archy [21]

Answer:

Step-by-step explanation:

No. of days Delynn played soccer = 3/5 of 45

= (3/5) * 45 = 3*45/5 = 3*9

= 27 days

8 0

3 years ago

Read 2 more answers

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago

Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)

timama [110]

<h3>Answer: c = 7/4</h3>

================================================

Work Shown:

Compute the function value at the endpoints

$f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\$

With a = -5 and b = 4, we have

$f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\$

So,

$f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\$

Use algebra to solve for c

$-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\$

6 0

3 years ago