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DanielleElmas [232]

3 years ago

13

Alice invests $8000 in two certificates of deposit; the interest on the first certificate is 2%, and the interest on the second

is 3%. Let's denote the amount she invests in the first certificate by x. Then the amount she invests in the second certificate is _________ . The interest she receives on the first certificate is __________ , and the interest she receives on the second certificate is ______________ . So a function that models the total interest she receives from both certificates is y =___________ .

1 answer:

Nataliya [291]3 years ago

7 0

1. 8000-x (You invest x in the first, then you have 8000-x more to make up for)
2. 0.02x (Interest * how much you invested)
3. 0.03(8000-x) (Interest * how much you invested)
4. 0.02x+240-0.03x
=240-0.01x

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The ordinate is twice the abscissa

zepelin [54]

We can construct the function:

$f(x)=2x$

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The function actually represents a line. Since it can have linear form.

$f(x)=2x+0$

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r3t40

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2 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:

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

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

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2 years ago

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vovikov84 [41]

A = {1, 3, 5, 7, 9}
B = {2, 4, 6, 8, 10}
C = {1, 5, 6, 7, 9}

(B ∪ C) = {1, 2, 4, 5, 6, 7, 8, 9, 10}
so
A ∩ (B ∪ C) = {1, 5, 7 , 9}

6 0

3 years ago

28) Find the circumference of a circle with a diameter of 2.5 inches. (it = 3.14)

antoniya [11.8K]

Answer: Circumference = 7.85 inches

Concept:

Here, we need to know the idea of circumference.

The circumference is the perimeter of a circle. The perimeter is the curve length around any closed figure.

Circumference = 2πr

r = radius

π = constant

Solve:

<u>Given information</u>

r = (1/2) d = (1/2) (2.5) = 1.25 inches

π = 3.14

<u>Given formula</u>

Circumference = 2πr

<u>Substitute values into the formula</u>

Circumference = 2 · (3.14) · (1.25)

<u>Simplify by multiplication</u>

Circumference = (2.5) · (2.14)

Circumference = $\boxed {7.85 inches}$

Hope this helps!! :)

Please let me know if you have any questions

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2 years ago

Read 2 more answers

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andriy [413]

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3 years ago