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3 years ago

7

The ratio of girls to boys competing at a gymnastics competition was 2 to 1 there were 75 kids participating in the competition.

How many boys were in this competition?

1 answer:

skad [1K]3 years ago

8 0

Answer:

25 boys

Step-by-step explanation:

1 of 3 kids was a boy, so (1/3)·75 = 25 were boys.

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Which of the following is equivalent to log 7^ 50 rounded to three decimal places?

otez555 [7]

7^50
= 1.798 × 10^42

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3 years ago

3x-7=4x+10 <br> show the steps pleeease

Fudgin [204]

Answer:

-17

Step-by-step explanation:

3x-7=4x+10

Add 7 to both sides & simplify

3x-7=4x+10

+7 +7

3x=4x+17

Subtract 4x from both sides & simplify

3x=4x+17

-4x -4x

-x=17

*Negative x equals -1*

Divide both sides by -1 & simplify

$\frac{-x}{-1} =\frac{-17}{-1}$

$x=-17$

3 0

3 years ago

Read 2 more answers

Which expression is equivalent to 8x-2x+x+x<br><br> a. 4x<br> b. 8x<br> c. 6x-2x<br> d. 10x-2y

Deffense [45]

B - 8x

To find this, combine like terms.

8x - 2x is 6x, then add the two x's on the side to bring it back to 8x.

Hope this helps!

4 0

3 years ago

Read 2 more answers

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

Al simplificar la fraccion 150/600 a su mínima expresión se obtiene 15/6<br>

zaharov [31]

Answer:

cierto

Step-by-step explanation:

8 0

3 years ago