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Naddika [18.5K]

3 years ago

14

What is the volume of this rectangular prism?

2 answers:

drek231 [11]3 years ago

7 0

Answer:

0.75 is the volume

Step-by-step explanation:

First, multiply.

1/2 × 1/2 = 0.25 or 1/4

Then, multiply again.

0.25 × 3 = 0.75

Have a great day! :)

lisov135 [29]3 years ago

4 0

Answer: I doubt your in it anymore

Step-by-step explanation:

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2 years ago

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Three of the angles of quadrilateral are 100 degrees, 104 degrees and 44 degrees. What is the size of the 4th angle

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All quadrilaterals should have 360 degrees.

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From a to D which graph represents a function ? And why ?

tatiyna

Option D I believe because it is the only one that passes the vertical line test (the test that for every X value, there is only 1 Y value).

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Mr. And Mrs. Kingsley went out for dinner on Valentine's Day. Their dinner bill came to $57.24. If they wish to leave a 15% tip

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Step-by-step explanation:

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3 years ago

F(x) = x^2+x-2/x^2-3x-4

Andrej [43]

i. Domain and Range

The given function is

$f(x)=\frac{x^2+x-2}{x^2-3x-4}$

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

$\frac{x^2+x-2}{x^2-3x-4}=0$

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

$f(0)=\frac{0^2+0-2}{0^2-3(0)-4}$

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

$x^2-3x-4=0$

This implies that;

$x^2+x-4x-4=0$

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

8 0

2 years ago

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