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4 years ago

Given that BE=6, CE=6,and FE measures 5 morethan DE, find the lengthof DE.

Mathematics

1 answer:

katen-ka-za [31]4 years ago

4 0

Answer:

DE = 4 units.

Step-by-step explanation:

By the chord intersection theorem:

BE * CE = DE * FE

But FE = DE + 5 so:

BE * CE = DE * (DE + 5)

6 * 6 = DE^2 + 5DE

DE^2 + 5DE - 36 = 0

(DE + 9)(DE - 4) = 0

So DE = -9 or 4

It can't be negative so DE = 4.

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3 years ago

What is the solution for -10 -19<br> x -1<br> x < -1

Natali [406]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the equation of a circle with center of (4,2) passing through the point (7,2)

attashe74 [19]

Answer:

(x-4)^2 + (y-2)^2 = 3^2

(x-4)^2 + (y-2)^2 = 9 (simplified if needed)

Step-by-step explanation:

-Equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$ where the center is (h, k) and the radius is $r^2$ .

-Place the center and the point onto that equation:

$(7-4)^2+(2-2)^2=r^2$

-Then, you solve:

$(7-4)^2+(2-2)^2=r^2$

$(3)^2+(0)^2=r^2$

$9 +0=r^2$

$9=r^2$

$\sqrt{9} =\sqrt{r^2}$

$3=r$

-So, the result is:

$(x-4)^2+(y-2)^2=3^2$

$(x-4)^2+(y-2)^2=9$ (simplified if needed)

3 0

3 years ago

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Sav [38]

About
2884 people practice a sport/club
1133 people nap
618 people work
1751 people do homework
1442 people play games
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3 years ago

Does the given equation of line form parallel lines or not

malfutka [58]

$k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2$

$k:x+4y=10\ \ \ |-x\\\\4y=-x+10\ \ \ \ |:4\\\\y=-\dfrac{1}{4}x+\dfrac{10}{4}\to m_1=-\dfrac{1}{4}\\\\l:3x-2y=5\ \ \ \ |-3x\\\\-2y=-3x+5\ \ \ \ |:(-2)\\\\y=\dfrac{3}{2}x-\dfrac{5}{2}\to m_2=\dfrac{3}{2}$

$m_1\neq m_2\to\text{ no parallel}\\\\m_1\cdot m_2=-\dfrac{1}{4}\cdot\dfrac{3}{2}\neq-1\to\text{ no perpendicular}$

Answer: NOT PARALLEL

5 0

4 years ago

Given that BE=6, CE=6,and FE measures 5 morethan DE, find the lengthof DE.​

Given that BE=6, CE=6,and FE measures 5 morethan DE, find the lengthof DE.