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Gekata [30.6K]
3 years ago
6

Anyone please help me it is a very simple question:

Mathematics
1 answer:
leonid [27]3 years ago
4 0
Try this;

6£(x) + 6£(x+3)
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HELP WITH PART C‼️Any questions ask
saveliy_v [14]

Answer:

they are alike because they both include a variable and they both have their totals in currency form. They are different because they use different variables.

Step-by-step explanation:

6 0
3 years ago
5 (5x6)+(8x9)+6=? 2(5x8)+(8x28)+9
KonstantinChe [14]

Answer:

The left side (228)does not equal the right side(313) that means the given statement is false

Step-by-step explanation:

8 0
2 years ago
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Solve the following integral.<br><br> <img src="https://tex.z-dn.net/?f=%5Cint4x%5Ccos%282-3x%29dx" id="TexFormula1" title="\int
bagirrra123 [75]

Hi there!

\boxed{-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C}

To find the indefinite integral, we must integrate by parts.

Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":

u = 4x

du = 4

dv = cos(2 - 3x)

v = 1/3sin(2 - 3x)

Write into the format:

∫udv = uv - ∫vdu

Thus, utilize the solved for expressions above:

4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx

Simplify:

-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx

Integrate the integral:

∫4/3(sin(2 - 3x)dx

u = 2 - 3x

du = -3dx ⇒ -1/3du = dx

-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C

Combine:

-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C

7 0
2 years ago
Read 2 more answers
Solve: z-5=75<br><br> A:z=71<br> B:z=80<br> C:z=70<br> D:z=79
poizon [28]

B: z=80___________________________________

5 0
3 years ago
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Subtract 8 y^2 − 5 y + 7 from 2 y^2 + 7 y + 1 1 <br><br> The answer is: −6y ^2 +12y+4
lys-0071 [83]

Answer:

\large\boxed{-6y^2+12y+4}

Step-by-step explanation:

(2y^2+7y+11)-(8y^2-5y+7)\\\\=2y^2+7y+11-8y^2-(-5y)-7\\\\=2y^2+7y+11-8y^2+5y-7\qquad\text{combine like terms}\\\\=(2y^2-8y^2)+(7y+5y)+(11-7)\\\\=-6y^2+12y+4

4 0
3 years ago
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