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3 years ago

9

What is 8x+4/2 in simplest form ?

2 answers:

Marrrta [24]3 years ago

5 0

The simplest form is 8 x + 2.

Varvara68 [4.7K]3 years ago

5 0

8x+4/2 in simplest form is 8x+2

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The polynomial 24x3 − 54x2 + 44x − 99 is factored by grouping.

Snezhnost [94]

The answer should be
24x3+44x-54x2-99

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3 years ago

A cone has a base radius of 7 centimeters and a height of 21 centimeters.

Arada [10]

V = (1/3) x πr² x h = cm³

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3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

<u>case B)</u> $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

<u>case C)</u> $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

<u>case D)</u> $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

<u>the answer is</u>

$f(x)=(x-2)(x+2)$

6 0

3 years ago

Read 2 more answers

Analyze the diagram below and complete the instructions that follow.

Jlenok [28]

Step-by-step explanation:

$scale \: factor = \frac{7}{28} = \frac{1}{4} \\$

$ratio \:of \:surface \:areas\\=\frac{2[5\times 7 + 7\times 3+3\times 5]}{2[20\times 28+ 28\times 12+12\times 20] }\\\\=\frac{[35 + 21+15]}{[560+ 336+240] }\\\\=\frac{71}{1136}\\\\=\frac{1}{16}\\\\= 1\: :\: 16$

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3 years ago

Y=-3/4(4/1)+1 i dont know how to solve it but I know it

olga nikolaevna [1]

Answer:

Slope: 0

Y-Intercept: −2

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3 years ago