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2 years ago

How do you write this is word form?

Mathematics

2 answers:

artcher [175]2 years ago

7 0

The word thing is do you 1010 million 310 million who 763,000 1136 storage go right this nuess

Vikki [24]2 years ago

5 0

Three hundred ten million seven hundred sixty-three thousand one hundred thirty-six hope this help

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Describe the transformation f(x) = 2x + 1 -> g(x) = -8(x + 4) - 1

katrin2010 [14]

f(x) = 2x + 1; Paren graph = x;

The graph was shifted one unit up and stretched 2 times.

g(x) = -8(x+4) - 1 = -8x - 32 - 1 = -8x - 33

The graph was shifted 33 units down and stretched 8 times and flipped 180 degrees.

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3 years ago

Ronnie charged his customers a flat fee of $25 to fix computers computers.Then he charge $30 for each hour that he spent fixing

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Step-by-step explanation:

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3 years ago

Is 3b-9 a algebraic expression

Slav-nsk [51]

Answer:

yes

Step-by-step explanation:

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3 years ago

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Graph the line y=kx+1 if it is known that the point M belongs to it: m(1,3)

Novosadov [1.4K]

Answer:

y = 2x +1

Step-by-step explanation:

y=kx+1

Substitute the point into the equation

3 = k(1) + 1

3 = k+1

Subtract 1 from each side

3-1 = k

2 = k

y = 2x +1

The slope is 2 and the y intercept is 1

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2 years ago

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Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$

Answer is choice D

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2 years ago

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