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2 years ago

Simplify the expression -6(2m+3)

Mathematics

2 answers:

padilas [110]2 years ago

6 0

The answer is -12m-18

Agata [3.3K]2 years ago

4 0

Answer:

-12m-18

Step-by-step explanation:

hope this helps!

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What is the domain and range of the function graphed below?

Nataliya [291]

Answer:

I think the answer is C.

Step-by-step explanation:

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3 years ago

What is the y-intercept of y= -3x

dybincka [34]

Answer:

(0,0) is the y intercept

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3 years ago

Read 2 more answers

Consider the initial value problem yâ€²+2y=4t,y(0)=8.

Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

$y'+2y=4t\\\\y(0)=8$

a) The algebraic equation obtain by using the Laplace transform is:

$L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\$

next, you replace (1) and (2):

$sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}$ (this is the algebraic equation)

$sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}$ (this is the solution for Y(s))

$y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}$

To find the inverse Laplace transform of the first term you use partial fractions:

$\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}$

Thus, you have:

$y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}$

(this is the solution to the differential equation)

5 0

3 years ago

For which pair of functions is (gºf)(a) =la|-2

nikklg [1K]

Answer:

<h2> The third</h2>

Step-by-step explanation:

$(g\circ f)(x) = g\big(f(x)\big)$

so:

$I.\quad (g\circ f)(a)=\sqrt{f(a)}=\sqrt{a^2-4}=\sqrt{(a+2)(a-2)}\\\\II.\quad (g\circ f)(a)=2f(a)-2=2(\frac12a-1)-2=a-2-2=a-4\\\\ \bold{III.\quad (g\circ f)(a)=\sqrt{f(a)-5}-2=\sqrt{5+a^2-5}-2=\sqrt{a^2}-2=|a|-2}\\\\ IV.\quad (g\circ f)(a)=4f(a)-5=4(3-3a)-5 = 12-12a-5=-12a+7$

8 0

3 years ago

Translate three more than number is eight

alexira [117]

Answer:

3+x=8

or this 3 < x

<h2 />

3 0

3 years ago