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Hunter-Best [27]
3 years ago
5

-5 (x-1)-x+1 = -18 help please​

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
4 0

Answer:

The value of x is 4.

Step-by-step explanation:

-5(x-1)-x+1= -18

or, -5x+5-x+1=-18

or, -6x+6= -18

or, -6(x-1)=-18

or, x-1=3

or, x=4

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Bumek [7]

Answer:

B) False.

Step-by-step explanation: 8 to the power of 2 plus 10 to the power of 2 gives you 164 but 18 to the power of 2 equals 324. In order to make a right triangle the answers have to be the same, but they are not so Paul cannot make a right triangle. Hope this answered your question.

7 0
3 years ago
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Please help!! 4 and 7 is a wrong answer.
Marta_Voda [28]

the answer is A but im just guessing ya know?

4 0
3 years ago
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Solve for b2 in A = 1/2 h ( b1 + b2 ), if A = 16, h = 4, and b1 = 3.
Katarina [22]

Hello from MrBillDoesMath!

Answer:

b2 = 5  


Discussion:

A = 1/2 h (b1 + b2) .

Substituting A = 16, h = 4, and b1=3 in the above formula gives:

16 = (1/2) (4)( 3 + b2)    =>         (as (1/2)4 = 2) )

16 = 2 ( 3 + b2)             =>         (divide both sides by 2)

8 = (3 + b2)                   =>         (subtract 3 from both sides)

8-3  = b2                       =>

5     =  b2


Check Area formula:

Does A = 16 = (1/2)(4)(3+5)   ?

Does 16 = (1/2) (4)(8)             ?

Does 16 = (1/2)(32)                ?  Yes it does so our calculation for b2 is correct




Thank you,

MrB

7 0
3 years ago
Classify the statement as either true or false.<br> lim<br> 6 = 6
asambeis [7]

Answer:

false

Step-by-step explanation:

if is 6= you need to time by it self like: 6×6=is ur answer

4 0
3 years ago
For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
3 years ago
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