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3 years ago

If bc=12 ab= 16 and ae =8 find ef

Mathematics

1 answer:

aev [14]3 years ago

3 0

Answer:

EF=6

Step-by-step explanation:

The complete question is shown in the attachment.

Triangle AEF is similar to triangle ABC.

This means that, the corresponding sides are proportional.

$\frac{EF}{BC}=\frac{EB}{AB}$

From the diagram, EB=16-8=8

AB=16

AE=8

BC=12

We substitute the values into the formula to get:

$\frac{EF}{12}=\frac{8}{16}$

Multiply both sides by 12 to get:

$12*\frac{EF}{12}=\frac{8}{16} *12$

$EF=\frac{1}{2}*12=6$

Therefore EF is 6 units

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Which is an equation of the parabola that intersects the x-axis at points (-1, 0) and (3, 0)? Show your work or explain how you

hichkok12 [17]

Answer:

$\displaystyle \large{y=x^2-2x-3}$

Step-by-step explanation:

The given points are (-1,0) and (3,0) to find an equation of parabola. Notice that they say or the given points are x-intercepts, which means they make it easier for us to find an equation of parabola using these x-intercepts.

Quick Note:

x-intercepts are simply the roots or solutions of quadratic equation, so if our root is x = -2 then we can write as (-2,0) and write back to x+2=0.

So what we are going to do is to write the roots’ equation back as $\displaystyle \large{x+\alpha =0}$

Our points or roots are given, therefore:

If x = -1 then x+1=0

If x = 3 then x-3=0

Then multiply x+1 with x-3 because if (x-1)(x+3)=0 then x-1=0 or x+3=0:

$\displaystyle \large{(x+1)(x-3)=0}$

Convert 0 to y since we want to find a function:

$\displaystyle \large{(x+1)(x-3)=y}\\\displaystyle \large{y=(x+1)(x-3)}$

Expand the expressions in and simplify to standard form:

$\displaystyle \large{y=x^2-3x+x-3}\\\displaystyle \large{y=x^2-2x-3}$

Therefore, the equation is $\displaystyle \large{y=x^2-2x-3}$

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Summary

If we are given the points $\displaystyle \large{(\alpha ,0)}$ and $\displaystyle \large{(\beta ,0)}$ with wideness of the graph or a-value = 1 then the parabola equation is:

$\displaystyle \large{y=(x-\alpha )(x-\beta )}\\\displaystyle \large{y=x^2-\beta x-\alpha x+\alpha \beta }\\\displaystyle \large{y=x^2-(\beta +\alpha )x+\alpha \beta }$