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3 years ago

PLEASE HELP!! Which graph represents the function y=sin(1/2x) after a shift of 3 units down?

Mathematics

2 answers:

Anastasy [175]3 years ago

8 0

Answer: A

Step-by-step explanation:

Got it right on edg

dybincka [34]3 years ago

3 0

Answer:

Step-by-step explanation:

correct on edge

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Reduce 6/12 to its simplest form

Vlad1618 [11]

Answer:

1/2

Step-by-step explanation:

6/12

3/6

1/2

7 0

3 years ago

Read 2 more answers

Hi. I need help with these See image for question Answer no 8 ,9 and 10

Nezavi [6.7K]

Answer:

8) 4 + 2q²/p² - 4r/p + r²/p²
9) (3/4, -9/4)
10) (3/8, 41/16)

Step-by-step explanation:

8. ============

Given

α and β are roots of px² + qx + r = 0

The sum of the roots is α + β = -q/p, the product of then roots αβ = r/p

(2 + α²)(2 + β²) =
4 + 2(α² + β²) + (αβ)² =
4 + 2((α + β)² -2αβ) + (αβ)² =
4 + 2((-q/p)² - 2r/p) + (r/p)² =
4 + 2q²/p² - 4r/p + r²/p²

------------------------------

9. ============

Given function

y = 2x² - 3x - 1

The minimum point is reached at vertex

The vertex is:

x = -b/2a
x = -(-3)/2*2 = 3/4

The corresponding y-coordinate is:

y = 2(3/4)² - 3(3/4) - 1 = 9/8 - 9/4 - 1 = 1/8(9 - 18 - 9) = - 18/8 = - 9/4

So the point is:

(3/4, -9/4)

---------------

10. ============

Given function

y = 2 - 3x - 4x²

The maximum is reached at vertex

The vertex is:

x = -b/2a
x = -(-3)/2(-4) = -3/8

The corresponding y-coordinate is:

y = 2 - 3(-3/8) -4(-3/8)² = 2 + 9/8 - 9/16 = 1/16(32 + 18 - 9) = 41/16

So the maximum point is:

(3/8, 41/16)

4 0

3 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$