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Ivenika [448]
3 years ago
8

Determine the axis of symmetry of a parabola whose graph is given below. Give your answer in the form X equals H

Mathematics
1 answer:
sergey [27]3 years ago
3 0

Answer:

x = -3

Step-by-step explanation:

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Jayson's membership card gives him $2.50 off his daily lunch salad. He paid $32.25 for 5 days. How much was the price of the sal
Vera_Pavlovna [14]

Answer:

44.75

Step-by-step explanation:

2.50×5=12.50.

12.50+32.25=44.75.

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What are all the roots of the function f(x)= x^3+3x^2-x-3
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Answer:

\mathrm{Domain\:of\:}\:x^3+3x^2-x-3\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}

Step-by-step explanation:

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2 years ago
What will be the volume of the box?
kolezko [41]

Answer:

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7 * 7 * 3

= 147 cm^3.

3 0
3 years ago
For which data sets is the median a better measure of center than the mean?
zheka24 [161]
A b d   those are the answers 

6 0
4 years ago
I need to solve this using l'hopital's rule and logarithmic diferentiation.
arlik [135]

Yo sup??

For our convenience let h=x+1

therefore

when x tends to -1, h tends to 0

hence we can rewrite it as

\lim_{h \to \ 0 } (cos(h))^{(cot(h^2 )}

This inequality is of the form 1∞

We will now apply the formula

e^(^g^(^x^)^(^f^(^x^)^-^1^)^)

plugging in the values of g(x) and f(x)

e^{lim_{h \to \ 0}{(cot(h)^2(cos(h)-1))}

express coth² as cosh²/sinh² and also write cosh-1 as 2sin²(h/2)

(by applying the property that cos2x=1-sin²x)

After this multiply the numerator and denominator with h² so that we can apply the property that

\lim_{x \to \ 0 } sinx/x =1

Now your equation will look like this.

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2)*h^2)/(sin(h)^2*h^2)}

We will now apply the result

\lim_{x \to \ 0 } sinx/x =1

where x=h²

we get

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2)}

we now multiply the numerator and denominator with 4 so that we can say

\lim_{h^2 \to \ 0 } sin^2(h/2)/(h^2/4) = 1

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2*4/4)}

=e^{lim_{h \to \ 0}{((cos(h)^2*2)/(4)}

Apply the limits and you will get

e^{cos(0)^2*2/4

=e^{1/2}

Hope this helps.

7 0
3 years ago
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