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3 years ago

Explain why 2,400 and 2,000 are both reasonable

Mathematics

2 answers:

Slav-nsk [51]3 years ago

5 0

Reasonable what? Estimates? Quotients? Products? We need more info to answer your question.

mina [271]3 years ago

3 0

You have to tell us reasonable for what????????? So we can answer your question! Maybe take a picture of the question and send it to me I'll answer it with what I think

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21+ 6m=3(m-1) not sure what to do here

castortr0y [4]

Answer:

M=-8

Step-by-step explanation:

21+6m=3(m−1)

Step 1: Simplify both sides of the equation.

21+6m=3(m−1)

21+6m=(3)(m)+(3)(−1)(Distribute)

21+6m=3m+−3

6m+21=3m−3

Step 2: Subtract 3m from both sides.

6m+21−3m=3m−3−3m

3m+21=−3

Step 3: Subtract 21 from both sides.

3m+21−21=−3−21

3m=−24

Step 4: Divide both sides by 3.

3m/3=-24/3

M=-8

5 0

3 years ago

Read 2 more answers

Hey peeps! Hows your day! Mine has been good/bad :)/:(

Debora [2.8K]

Answer:

Mine has been good so far

Step-by-step explanation:

6 0

3 years ago

Graph the solution of this inequality:<br> 4.52 – 100 > 125<br> Drag a point to the number line.

creativ13 [48]

Answer:

starting on 50 going right, using the open circle

Step-by-step explanation:

4 0

3 years ago

Add (9x + 6) + (2x - 9)

otez555 [7]

Answer:

11x-3

Step-by-step explanation:

open parenthesis

9x+6+2x-9

add like terms

11x-3

8 0

3 years ago

Find the horizontal asymptote of f of x equals quantity negative 2 x squared plus 3x plus 6 end quantity over quantity x squared

GREYUIT [131]

Answer:

y = -2 (none of the options)

Step-by-step explanation:

Horizontal Asymptotes

A function f is said to have a horizontal asymptote y=a, if

$\displaystyle \lim _{x\rightarrow -\infty }f(x)=a$

Or,

$\displaystyle \lim _{x\rightarrow +\infty }f(x)=a$

The function provided in the question is

$\displaystyle f(x)=\frac{-2x^2+3x+6}{x^2+1}$

We compute the limit

$\displaystyle \lim _{x\rightarrow \infty }\frac{-2x^2+3x+6}{x^2+1}$

Dividing by $x^2$

$\displaystyle \lim _{x\rightarrow \infty }\frac{-2+3/x+6/x^2}{1+1/x^2}$

Knowing that

$\displaystyle \lim _{x\rightarrow \infty }\frac{K}{x^n}=0,\ n\geqslant 1$

The limit results

$\displaystyle \lim _{x\rightarrow \infty }\frac{-2+0+0}{1+0}=-2$

The only horizontal asymptote is $y=-2$

None of the options presented is correct

4 0

3 years ago