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enyata [817]
3 years ago
15

RIGHT ANSWER WILL BE MARK BRAINLIEST

Mathematics
2 answers:
nirvana33 [79]3 years ago
6 0
...............the answer is D
kkurt [141]3 years ago
4 0
C is the correct answer.
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The answer is: x is choose your answer
katrin2010 [14]

Answer:

All real numbers

Let's find the critical points of the inequality.

−4x+7=17

−4x+7+−7=17+−7(Add -7 to both sides)

−4x=10

−4x−1=10−1

(Divide both sides by -1)

4x=−10

4x=−10(Solve Exponent)

log(4x)=log(−10)(Take log of both sides)

x*(log(4))=log(−10)

x=log(−10)log(4)

x=NaN

3 0
4 years ago
PLEASE CAN ANYONE HWLP WITH THIS QUESTION I NEED HELP ASAP!!!!!!
matrenka [14]

Answer: C.) $51

Step-by-step explanation:

$85.00 x 40%

85.00 x 0.40= 34

85.00-34=51.00

$51

3 0
3 years ago
Read 2 more answers
How do you do this question?
vaieri [72.5K]

Answer:

(-∞, -2), (-2, -0.618), and (1.618, 3)

Step-by-step explanation:

The red plus (+) signs indicate the regions in which the function is concave up, and the red negative (-) signs indicate the regions in which the function is concave down.

Note that the sign of the concavity changes at an inflection point.

Let's examine the intervals given.

(-∞, -2): Yes, concave up.

(-∞, -1.17): No. Concave up in (-∞, -2) but concave down in (-2, -1.17).

(-2, 0): No. Concave down in (-2, -1.17) but increasing in (-1.17, 0.0).

(-1.17, 0.689): Yes. Concave up.

(-0.618, 1.618): No. Concave up in (-0.618, 0.689) but concave down in (0.689, 1.618).

(0, 3): No. Concave up in (0, 0.689), concave down in (0.689, 2.481), and concave up in (2.481, 3).

(2.481, ∞): Yes. Concave up.

The three intervals that are concave up are (-∞, -2), (-1.17, -0.689), and (2.481, ∞).

4 0
3 years ago
The volume inside of a sphere is V=4πr33 where r is the radius of the sphere. Your group has been asked to rearrange the formula
Slav-nsk [51]

Answer:

Step-by-step explanation:5

8 0
3 years ago
Inverse function. sketch the graph of each function.​
Nikitich [7]

Answer:

I cant see picture

Step-by-step explanation:

8 0
3 years ago
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