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1 year ago

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

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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