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3 years ago

8

Find the area of the figure. (sides meets at right angle)

1 answer:

S_A_V [24]3 years ago

3 0

Multiply length time width, and make sure you square your answer!

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In the algebraic expression 5x+8y-12z, the values 5, 8, and -12 are

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3 years ago

Martha ran 2 miles in 15 minutes.<br> At this rate, how many miles<br> would she run in 2 hours?

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3 years ago

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3 years ago

The graph of which function has an axis of symmetry at x =-1/4 ?

Phoenix [80]

we know that

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

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

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

so

$x=h$ ------> axis of symmetry of a vertical parabola

we will determine in each case the axis of symmetry to determine the solution

<u>case A)</u> $f(x)=2x^{2}+x-1$

<u>Convert to vertex form</u>

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

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

Factor the leading coefficient

$f(x)+1=2(x^{2}+0.5x)$

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

$f(x)+1+0.125=2(x^{2}+0.5x+0.0625)$

$f(x)+1.125=2(x^{2}+0.5x+0.0625)$

Rewrite as perfect squares

$f(x)+1.125=2(x+0.25)^{2}$

$f(x)=2(x+0.25)^{2}-1.125$

the vertex is the point $(-0.25,-1.125)$

the axis of symmetry is

$x=-0.25=-\frac{1}{4}$

therefore

the function $f(x)=2x^{2}+x-1$ has an axis of symmetry at $x=-\frac{1}{4}$

<u>case B)</u> $f(x)=2x^{2}-x+1$

<u>Convert to vertex form</u>

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

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

Factor the leading coefficient

$f(x)-1=2(x^{2}-0.5x)$

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

$f(x)-1+0.125=2(x^{2}-0.5x+0.0625)$

$f(x)-0.875=2(x^{2}-0.5x+0.0625)$

Rewrite as perfect squares

$f(x)-0.875=2(x-0.25)^{2}$

$f(x)=2(x-0.25)^{2}+0.875$

the vertex is the point $(0.25,0.875)$

the axis of symmetry is

$x=0.25=\frac{1}{4}$

therefore

the function $f(x)=2x^{2}-x+1$ does not have a symmetry axis in $x=-\frac{1}{4}$

<u>case C)</u> $f(x)=x^{2}+2x-1$

<u>Convert to vertex form</u>

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

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

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

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

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

Rewrite as perfect squares

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

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

the vertex is the point $(-1,-2)$

the axis of symmetry is

$x=-1$

therefore

the function $f(x)=x^{2}+2x-1$ does not have a symmetry axis in $x=-\frac{1}{4}$

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

<u>Convert to vertex form</u>

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

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

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

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

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

Rewrite as perfect squares

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

the vertex is the point $(1,0)$

the axis of symmetry is

$x=1$

therefore

the function $f(x)=x^{2}-2x+1$ does not have a symmetry axis in $x=-\frac{1}{4}$

<u>the answer is</u>

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

7 0

4 years ago

Read 2 more answers

Can someone help me answer this

rosijanka [135]

Add up the times spent

8 0

4 years ago

Read 2 more answers