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3 years ago

HELP!!! What is the missing term , x, in the pattern?

Mathematics

1 answer:

maxonik [38]3 years ago

8 0

the missing value will be -36 because if you multiplied it with 4 you will only have to add another -4 to get -144

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Can someone help me with this

VikaD [51]

Answer:

A. y = 3x.

Step-by-step explanation:Relation 2 is a straight line that does not pass through the origin, so it is a partial variation.

6 0

2 years ago

Find the general indefinite integral. (use c for the constant of integration. )

Vlada [557]

The indefinite integral will be $\int (7x^2+8x-2)dx=\dfrac{7x^3}{3}+4x^2-2x+C)$

<h3 /><h3>what is indefinite integral?</h3>

When we integrate any function without the limits then it will be an indefinite integral.

General Formulas and Concepts:

Integration Rule [Reverse Power Rule]:

$\int x^ndx=\dfrac{x^{n+1}}{n+1}}+C$

Integration Property [Multiplied Constant]:

$\int cf(x)dx=c\int f(x)dx$

Integration Property [Addition/Subtraction]:

$\int [f(x)\pmg(x)]dx=\int f(x)dx\pm \intg(x)dx$

[Integral] Rewrite [Integration Property - Addition/Subtraction]:

$\int (7x^2+8x-2)dx=\int 7x^2dx+\int 8xdx -\int 2dx$

[Integrals] Rewrite [Integration Property - Multiplied Constant]:

$\int (7x^2+8x-2)dx=7 \int x^2dx+ 8 \int xdx -2\int dx$

[Integrals] Reverse Power Rule:

$\int (7x^2+8x-2)dx= 7(\dfrac{x^3}{3})+8(\dfrac{x^2}{2})-2x+C$

Simplify:

$\int (7x^2+8x-2)dx= \dfrac{7x^3}{3}+4x^2-2x+C$

So the indefinite integral will be $\int (7x^2+8x-2)dx= \dfrac{7x^3}{3}+4x^2-2x+C$

To know more about indefinite integral follow

brainly.com/question/27419605

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7 0

2 years ago

35 - 10 ➗ 5 + [(5 + 3) • 4]

snow_lady [41]

<span>35 - 10 ➗ 5 + [(5 + 3) • 4]
Do PEMDAS
1.52</span>

7 0

3 years ago

Read 2 more answers

Hey I see you there so wanna help a girl out ?

netineya [11]

Answer:

ITS probably like 70 or something

Step-by-step explanation:

3 0

2 years ago

Q. 9 [Decimals] (10 Points)<br> Write the following as a decimal:<br> nine tenths

never [62]

It would be zero point 9 or 0.9

7 0

2 years ago