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uranmaximum [27]
3 years ago
5

3 x + 9 = box If the equation has no solution, which expression can be written in the box on the other side of the equation?

Mathematics
1 answer:
hichkok12 [17]3 years ago
7 0

Answer:

3(x+4)

Step-by-step explanation:

Cause I just toke the test on e d g e n u !

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drag  10 to the

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What is a cube in math
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In arithmetic and algebra, the cube of a number n is its third power, that is, the result of multiplying three instances of n together. The cube is also the number multiplied by its square: n3 = n × n2 = n × n × n.

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In August, Emily's Clothing Store sold 98 long sleeve shirts with the ratio of short sleeve to long sleeve being 3:7. How many s
Gnoma [55]

Answer:

the number of short sleeves sold is 42

Step-by-step explanation:

The computation of the number of short sleeves sold is shown below:

Given that

The clothing store sold 98 long sleeve shirts

And, the ratio of short sleeve to long sleeve is 3:7

So, the number of short sleeves sold is

= 98 × 3 ÷ 7

= 42

Hence, the number of short sleeves sold is 42

6 0
3 years ago
Students did push-ups for a fitness test. Their goal was 20 push-ups. For each student,
ale4655 [162]

Answer:

1- 80%

2- 210%

3- 65%

4- 105%

Step-by-step explanation:

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
4 years ago
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