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3 years ago

Will give Brainly lollllllll XD......................

Mathematics

2 answers:

Vladimir79 [104]3 years ago

8 0

Answer:

-1

Step-by-step explanation:

marshall27 [118]3 years ago

6 0

Answer:

it looks like it would just be -1

Step-by-step explanation:

there two -x and two positive x which would cancel out. three positive 1 and four negative 1. -4+3= -1

that's the best I could come up with

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The domain of a function f(x) is the set of all integers from 0 to 10, and the range of f(x) is the set of all in

e-lub [12.9K]

I think that the correct answer is f(-1)=-8 and f(0)=-5
I hope this helps :D

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3 years ago

Find f(-5) if f( x ) = | x + 1|.

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

f(x)=x+1

then,

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=-4

8 0

2 years ago

1 1/16 cups of butter times 20 equals how much

iren2701 [21]

13.75 Cups in us cups

8 0

3 years ago

Read 2 more answers

Andy estimated that he would need 75 feet of lumber for a tree house project. he later found that the actual amount of lumber ne

Kipish [7]

Step 1
Calculate the error
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step 2
Divide the error by the exact value
11/64=0.1719

step 3
Convert that to a percentage
0.1719*100=17.19%

the answer is
17.19%

8 0

3 years ago

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