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podryga [215]
2 years ago
5

Find the net price if the $10.00 list price is discounted 10%.

Mathematics
2 answers:
yanalaym [24]2 years ago
4 0

Answer:

9.00

Step-by-step explanation:

MatroZZZ [7]2 years ago
3 0

Answer:

the new price would be $9.00

Step-by-step explanation:

%10 of 10 is 1

10-1=9

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C.the height of a student and the the day of a month a student was born
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1<br> Solve - (8x – 12) = 2x – 3<br> 4<br> 218x -<br> pleaseee help!!
Alex Ar [27]

Answer:

1/4.(-96)=2x-3

-1/4.96=2x-3

-24=2x-3

-2x-24=-3

-2x=-3+24

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x=-21/2

Step-by-step explanation:

8 0
2 years ago
2 = (-x) Help I'm a not so bright blonde!
makvit [3.9K]
If you need to solve this equation you can multiply the both side of equation on (-1). Like that
<span>2 = (-x);
-1·2 = (-1)·(-x)
if you multiply negative and </span><span>positive </span>you'll have negative,
if you multiply negative and <span>negative you'll have </span><span><span>positive. So
</span> 
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5 0
3 years ago
PLZ ANSWER ASAP! 15 points
Mariulka [41]

Answer: THe last one.

Step-by-step explanation:

This question is, in essence, basically asking which numbers are less than -3. Looking at the numbers, it is clearly the last set.

5 0
3 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
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