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3 years ago

If a/b and c/d are rational expressions then a/b divided by c/d =a times d/b times c true or false

Mathematics

1 answer:

Tomtit [17]3 years ago

4 0

Answer:

(a d)/(bc)

Step-by-step explanation:

a/b ÷ c/d

Copy dot flip

a/b * d/c

ad / bc

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Determine whether the ratios 35 feet every 15 seconds and 7 feet every 2 seconds are equivalent.

murzikaleks [220]

Answer:

Step-by-step explanation:

Find the rate of those two statements:

28 ÷ 16 = 1.75

1.75 ft per seconds

7 ÷ 4 = 1.75

1.75 ft per seconds

Two statements have the same rate, therefore 28 feet every 16 seconds and 7 feet every 4 seconds are equivalent

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1 year ago

Marie has renters insurance that she must pay twice a year. If each payment is $96, how much money should she set aside each mon

enyata [817]

$96(2 payments) = $192

$192/(12 months) = $16 per month (she should set aside)

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3 years ago

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What is the median of this set values?

Ierofanga [76]

The answer is D. 18.5
How? Because you have the numbers in order and cross the out it leaves you with 17, and 20 and the number between can be 19 or 18.5.

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3 years ago

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Complete each of these tables. What do you notice?

Rama09 [41]

Answer:

-4

Step-by-step explanation:

they have the same value for each couple

3 0

3 years ago

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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