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3 years ago

Which statement of the function is true ????

Mathematics

2 answers:

Lera25 [3.4K]3 years ago

3 0

B is the answer when the parabola is facing down it is negative

vovangra [49]3 years ago

3 0

Answer:

The function is negative for all real values of x where x < -3 and x > 1.

Step-by-step explanation:

The values of the function are the y-values. The y-values are positive from 3≤x≤1. Everywhere else in the function, the y-values are negative.

This means that the function is negative from x < -3 and from x > 1.

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7^14 ÷ 7^2<br> I-ready btw

Lyrx [107]

Answer:

7^12

Step-by-step explanation:

7^14

-----------

7^2

When we divide with the same base, we subtract the exponents

7^(14-2)

7^12

5 0

3 years ago

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A rock is dropped from the edge of a 550 foot cliff. It's position above the ground at any given time (t) is modeled by the func

Gemiola [76]

S(5) = -16*(5)*2 + 550 = 390

you just replace t with 5

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3 years ago

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What is 2 times the square root of 9 minus the square root of 4 expressed in simplified form?

Leokris [45]

Well I believe it's b

8 0

3 years ago

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Can you please solve 1 and 2 for me please? This doesn't even feel like math, just a stupid riddle.

kherson [118]

Answer:

1. Why does the graph not start at zero?

It doesn't start at zero because all ovens start at 350°(probably less, but normally 350°) and not zero degrees.

2. What does it mean when the graph flattens out?

When the graph flattens out, it means that the temperature doesn't increase nor decrease, but instead it remains the same.

6 0

3 years ago

Please help me with this question

san4es73 [151]

Step-by-step explanation:

Given: $f'(x) = x^2e^{2x^3}$ and $f(0) = 0$

We can solve for f(x) by writing

$\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx$

Let $u = 2x^3$

$\:\:\:\:du=6x^2dx$

Then

$\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du$

$\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k$

We know that f(0) = 0 so we can find the value for k:

$f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}$

Therefore,

$\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)$

5 0

3 years ago