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Scorpion4ik [409]

2 years ago

8

PLEASE HELP EMERGENCY

2 answers:

makvit [3.9K]2 years ago

6 0

Answer:

he because the Spence or bentualy

Troyanec [42]2 years ago

5 0

He because Spence butualy

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164 is 55% of what number?<br><br> Please somebody help me

monitta

This is how you solve it; but first of all convert the percent into a decimal;55%= 0.55

0.55x = 164 (divide by 0.55 on both sides)
x = 164/0.55
x = 298.18 (rounded to the nearest hundredth)
so 164 is 55% of 298.18(rounded)

8 0

3 years ago

the difference of two numbers is 16. The greater number is 5 less than 4 times the smaller number find the numbers

inna [77]

L-s=16
4s-5=l
4s-l=5
3s=21
s=7
l-7=16
l=16+7
l=23
The smaller number is 7 and the greater number is 23

5 0

3 years ago

Don't really understand how to solve please help <3

valina [46]

This is how the first one would be done. sorry if it’s a little messy the answer is 9. but try using photomath for the rest. it gives you the answer and explains how to do it

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3 years ago

Describe how to label a bar model

marishachu [46]

Wow how do you not know how and what kind of bar are you talking about

3 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