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nataly862011 [7]
3 years ago
15

If the side length of square ABCD is four times the side length of square PQRS, which statement is true? A. The perimeter of squ

are ABCD is 16 times the perimeter of square PQRS. B. The perimeter of square ABCD is 8 times the perimeter of square PQRS. C. The perimeter of square ABCD is 4 times the perimeter of square PQRS. D. The perimeter of square ABCD is 2 times the perimeter of square PQRS.
Mathematics
2 answers:
nataly862011 [7]3 years ago
7 0
C. The perimeter of square ABCD is 4 times the perimeter of square PQRS.
dolphi86 [110]3 years ago
4 0

Answer:

C. The perimeter of square ABCD is 4 times the perimeter of square PQRS.

Step-by-step explanation:

In order to calculate this we can just put values into the sides of the squares, for example, we know that the side of ABCD is 4 times the side of PQRS, so if the side of PQRS is 1 unit, the side of ABCD will be 4 units.

The perimeter is calculated by adding all the sides, if a square has 4 equal sides the perimeter will be side x 4.

Perimeter of PQRS= 1 unit* 4= 4

Perimeter od ABCD=4 units*4=16

So the perimeter of ABCD is four times the perimeter of square PQRS.

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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
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