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Alchen [17]
4 years ago
5

What is the length of segment AE?

Mathematics
1 answer:
nexus9112 [7]4 years ago
6 0

Answer:

AE=\frac{50}{3}\ units

Step-by-step explanation:

step 1

In the right triangle ABC

Applying the Pythagoras Theorem fin the hypotenuse AC

AC^{2} =AB^{2}+BC^{2}

substitute

AC^{2} =6^{2}+8^{2}

AC^{2} =100

AC =10\ units

step 2

we know that

If two figures are similar, then the ratio of its corresponding sides is equal

so

\frac{AB}{CD}=\frac{AC}{CE}

substitute and solve for CE

\frac{6}{4}=\frac{10}{CE}\\ \\CE=4*10/6\\ \\CE=\frac{20}{3}\ units

step 3

Find the length of segment AE

AE=AC+CE

substitute the values

AE=10\ units+\frac{20}{3}\ units=\frac{50}{3}\ units

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Refer to this previous solution set

brainly.com/question/26114608

===========================================================

Problem 4

Like the three earlier problems, we'll place the kicker at the origin and have her kick to the right. The two roots in this case are x = 0 and x = 20 to represent when the ball is on the ground.

This leads to the factors x and x-20 and the equation y = ax(x-20)

We'll plug in (x,y) = (10,28) which is the vertex point. The 10 is the midpoint of 0 and 20 mentioned earlier.

Let's solve for 'a'.

y = ax(x-20)\\\\28 = a*10(10-20)\\\\28 = -100a\\\\a = -\frac{28}{100}\\\\a = -\frac{7}{25}\\\\

This then leads us to:

y = ax(x-20)\\\\y = -\frac{7}{25}x(x-20)\\\\y = -\frac{7}{25}x*x-\frac{7}{25}x*(-20)\\\\y = -\frac{7}{25}x^2+\frac{28}{5}x\\\\

The equation is in the form y = ax^2+bx+c with a = -\frac{7}{25}, \ b = \frac{28}{5}, \ c = 0

The graph is below in blue.

===========================================================

Problem 5

The same set up applies as before.

This time we have the roots x = 0 and x = 100 to lead to the factors x and x-100. We have the equation y = ax(x-100)

We'll use the vertex point (50,12) to find 'a'.

y = ax(x-100)\\\\12 = a*50(50-100)\\\\12 = -2500a\\\\a = -\frac{12}{2500}\\\\a = -\frac{3}{625}\\\\

Then we can find the standard form

y = ax(x-100)\\\\y = -\frac{3}{625}x(x-100)\\\\y = -\frac{3}{625}x*x-\frac{3}{625}x*(-100)\\\\y = -\frac{3}{625}x^2+\frac{12}{25}x\\\\

The graph is below in red.

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