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tresset_1 [31]
1 year ago
14

Given the definitions of f(x) and g(x) below, find the value of f(g(0)).

Mathematics
1 answer:
spayn [35]1 year ago
3 0

Answer:

f(g(0)) = 45

Step-by-step explanation:

to evaluate f(g(0)) , evaluate g(0) then substitute the value obtained into f(x)

g(0) = - 3(0) - 3 = 0 - 3 = - 3 , then

f(- 3) = 2(- 3)² - 5(- 3) + 12

       = 2(9) + 15 + 12

       = 18 + 27

      = 45

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If an open box has a square base and a volume of 115 in.3 and is constructed from a tin sheet, find the dimensions of the box, a
Karolina [17]
Let h = height of the box,
x = side length of the base.

Volume of the box is  V=x^{2} h = 115. 
So h = \frac{115}{ x^{2} }

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
S = 2( x^{2}  + hx + hx)  \\ = 2 x^{2}  + 4hx  \\  = 2 x^{2}  + 4( \frac{115}{ x^{2} } )x
= 2 x^{2} + \frac{460}{x}
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.

S' = 4x- \frac{460}{ x^{2} }  = 0
4x = \frac{460}{ x^{2} }  \\  4x^{3} = 460  \\ x^{3} = 115  \\ x =  \sqrt[3]{115} = 4.86
Then h=  \frac{115}{4.86^{2}} = 4.87
So the box is 4.86 in. wide and 4.87 in. high. 

5 0
3 years ago
Read 2 more answers
Half a sphere is called a ?
Vesna [10]

Answer:

half a sphere is called a hemisphere

7 0
3 years ago
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What is the slope of a line that goes through the points (10, -2) and (5,-5)? m =
Vinvika [58]

Answer:

m = 3/5

Step-by-step explanation:

use the formula so m = -5 - -2 / 5 - 10 , so m = -5+2 / -5

m = -3/-5 so m = 3/5

6 0
3 years ago
Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie
Dvinal [7]
For part (a), you have

\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}
x=a(x-2)+b(x+3)

If x=2, then 2=b(2-3)\implies b=-2.

If x=-3, then -3=a(-3-2)\implies a=\dfrac35.

So,

\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}

In the remainder term, the denominator x^2+x+2 can't be factorized into linear components with real coefficients, since the discriminant is negative (1-4\times1\times2=-7). However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i
\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)
\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)

Then you have

\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}
x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)

When x=-\dfrac12-\dfrac{\sqrt7}2i, you have

-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)
\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib
b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)

When x=-\dfrac12+\dfrac{\sqrt7}2i, you have

-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)
\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia
a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)

So, you could write

\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}

but that may or may not be considered acceptable by that webpage.
5 0
3 years ago
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What is the y intercept of the quadratic function f(x)=x^2+x-6
saveliy_v [14]

Answer:

(0,\, -6).

Step-by-step explanation:

On a cartesian plane, the y-intercept of a function is the point where the graph of that function intersects with the y\!-axis.

The y-axis of a cartesian plane is the same as the equation x = 0 (that is, the collection of all points with an x-coordinate of 0.)

Construct a system of two equations, with one equation representing y-axis and y = f(x) to represent the graph of this function:

\begin{aligned}\begin{cases} y = x^{2} + x - 6 & \text{for the quadratic function} \\ x = 0 & \text{for the $y$-axis}\end{cases}\end{aligned}.

Solve this system for x and for y. If a solution exists, then the y\!-axis and the graph of y = f(x) would indeed intersect. The point (x,\, y) would be the intersection of the y\!\!-axis and the graph of y = f(x)\!.

Substitute the second equation of the system into the first.

\begin{aligned}\begin{cases} x = 0 \\ y = -6\end{cases}\end{aligned}.

Hence, the intersection of the y-axis and the graph of y = f(x) would be (0,\, -6). By definition, this point would be the y\!-intercept of y = f(x)\!.

4 0
2 years ago
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