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Kipish [7]
3 years ago
6

Please help me!!!!!!!!!!!

Mathematics
1 answer:
lbvjy [14]3 years ago
6 0

Answer:I think its C but im not sure

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For number 6, evaluate the definite integral.
maks197457 [2]
\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\

\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64

\bf \\\\
-------------------------------\\\\
\displaystyle  \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2}  \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3
3 0
3 years ago
What does y= 3x - 2 equal?<br><br><br>​
cestrela7 [59]
THE ANSWER WOULD BE X=2/3
8 0
3 years ago
Read 2 more answers
Choose the correct response​
lilavasa [31]

Answer:

f(2) = 9

Step-by-step explanation:

It's give in the question,

Two different functions are,

f(x) = 4x + 1

g(x) = -2x

Then we have to find the output value of function 'f' for the input value x = 2.

By substituting the value of x in the function 'f',

f(2) = 4(2) + 1

     = 9

f(2) = 9

5 0
2 years ago
How do I solve question 2?
sdas [7]

you multiply the number of years

6 0
3 years ago
Construct 3 linear equation starting with qiven solution z = 1/3
Andreyy89

Answer:

(a)9z+2=5

(b)21z-11=-4

(c)4z=2-2z

Step-by-step explanation:

We are required to construct 3 linear equations starting with the given solution z = 1/3.

<u>Equation 1</u>

<u />z=\frac{1}{3}<u />

Multiply both sides by 9

9z=\frac{1}{3}\times 9\\9z=3

Rewrite 3 as 5-2

9z=5-2

Add 2 to both sides

Our first equation is: 9z+2=5

<u>Equation 2</u>

<u />z=\frac{1}{3}<u />

Multiply both sides by 21

21z=\frac{1}{3}\times 21\\21z=7

Rewrite 7 as 11-4

21z=11-4

Subtract 11 from both sides

Our second equation is: 21z-11=-4

<u>Equation 3</u>

<u />z=\frac{1}{3}<u />

Multiply both sides by 6

6z=\frac{1}{3}\times 6\\6z=2

Rewrite 6z as 4z+2z

4z+2z=2

Subtract 2z from both sides

Our third equation is: 4z=2-2z

4 0
3 years ago
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