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3 years ago

Someone who can help would be great.

Mathematics

1 answer:

Ne4ueva [31]3 years ago

6 0

Answer:

x=5

Step-by-step explanation:

In a square, the diagonals are equal.

4x+17 = 12x-23

Subtract 4x from each side.

4x-4x+17 = 12x-4x-23

17 = 8x-23

Add 23 to each side

17+23 = 8x-23+23

40 = 8x

Divide each side by 8

40/8 = 8x/8

5 =x

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Malik Brooks works 40 hours per week, 52 weeks each year for Metro Delivery. He earns $9.15 per hour. If Malik receives a 4.9% m

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His new hourly rate will be $9.60

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3 years ago

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According to the Complex Conjugate Root Theorem, if a+bi is a root of a quadratic equation, then __blank__ is also a root of the

Trava [24]

Answer:

a-bi

Step-by-step explanation:

If a quadratic equation lx^2+mx+n=0 has one imaginary root as a+bi then the other root is the conjugate of a+bi = a-bi

Because we have l, m and n are real numbers and they are the coefficients.

Sum of roots = a+bi + second root = -m/l

When -m/l is real because the ratio of two real numbers, left side also has to be real.

Since bi is one imaginary term already there other root should have -bi in it so that the sum becomes real.

i.e. other root will be of the form c-bi for some real c.

Now product of roots = (a+bi)(c-bi) = n/l

Since right side is real, left side also must be real.

i.e.imaginary part =0

bi(a-c) =0

Or a =c

i.e. other root c-bi = a-bi

Hence proved.

3 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:

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

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 $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

In the diagram, segments AD and AB are tangent to circle C. Solve for x.

Irina-Kira [14]

The answer is A !!!!!!!!

the two segments AD and AB are equal therefore if you make the two equal to each other you will get the answer:

x^2+2=11 then you subtract two to put x squared on the side by itself
then you have x^2=9
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4 years ago

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The _____ _____, f(x) = x, is formed by the composition of a function and its inverse. (2 words)

jonny [76]

Answer:

Identity function

Step-by-step explanation:

We are given the following in the question:

Let f(x) be a function.

Then, inverse of f(x) can be written as:

$f^{-1}(x)$