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NNADVOKAT [17]
3 years ago
12

Eight less than a number is no more than 14

Mathematics
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

x - 8 ≤ 14

Step-by-step explanation:

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In 2005 13.1 out of every 50 employees at a company were women. If there are 41,330 total employees estimate the number of women
ZanzabumX [31]

Answer:

10829 women

Step-by-step explanation:

41330 x 13.1 / 50 = 10828.46, or about 10829 women

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3 years ago
ZABD and ZDBC are supplementary angles.
adell [148]

x=54°

Answer:

x+126=180 supplementary

x=180-126

x=54

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3 years ago
Mrs. Tinsley’s favorite smoothie consists of 2 cups ice, 3 cups strawberries, and 4 cups orange juice. a. Ratio of strawberries
mamaluj [8]
3 to 4 is the ratio of strawberries to orange juice.
please give brainliest and have a great day!
4 0
3 years ago
Read 2 more answers
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
Simplify the expression 5^3 x 5^-5
kotykmax [81]

Answer:

\frac{1}{5^2} --- 1 over 5 squared

Step-by-step explanation:

When multiplying terms with a common base, you just add the exponents:

x^a\times x^b=x^{a+ b}

That's true even when you don't have any exponents.

5\times5=5^1\times5^1=5^{1+1}=5^2=25

\rightarrow5^3\times5^{-5}\\\rightarrow5^{3-5}\\\rightarrow5^{-2}

A negative exponent isn't fully simplified, so there's another rule to use:

x^{-y}=\frac{1}{x^y}

That is '1 over x to the y' if it's too small to read.

\rightarrow5^{-2}=\frac{1}{5^2}

4 0
2 years ago
Read 2 more answers
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