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Zepler [3.9K]
3 years ago
6

AB and AD are tangent to circle C. Find the length of AB. A. 39 B. 16 C. 3 D. 13

Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

B. 16

Step-by-step explanation:

Since AB and AD are tangent to circle C, they equal each other.

2x + 7 = 3x - 9     Make them equal each other

2x + 7 = 3x - 9      Subtract 2x from both sides

7 = x - 9            Add 9 to both sides

16 =  x            

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Step-by-step explanation:

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What is the standard form of the line with x-intercept of 6 and y-intercept of 5?
Vanyuwa [196]

Answer:

It should be 5x-6y=30

Step-by-step explanation:

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3 years ago
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Please help!
crimeas [40]

Answer:

The zeros are:

x =4, x=2, x = 5

  • The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

0=\left(x-4\right)^2\left(x^2-7x+10\right)

switch sides

\left(x-4\right)^2\left(x^2-7x+10\right)=0

as

x^2-7x+\:10=\left(x-2\right)\left(x-5\right)

so

\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0

Using the zero factor principle

  • \mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x =4, x=2, x = 5

Thus, the zeros are:

x =4, x=2, x = 5

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

  • The function has three distinct real zeros.

Hence, option (B) is true.

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3 years ago
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69 square feet is the answer
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3 years ago
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<img src="https://tex.z-dn.net/?f=%20-%203%20%2B%205i%20%20%5Cdiv%20%20-%203%20-%204i" id="TexFormula1" title=" - 3 + 5i \div
Likurg_2 [28]

Answer:

\frac{-11}{25}+\frac{-27}{25}i given you are asked to simplify

\frac{-3+5i}{-3-4i}

Step-by-step explanation:

You have to multiply the numerator and denominator by the denominator's conjugate.

The conjugate of a+bi is a-bi.

When you multiply conjugates, you just have to multiply first and last.

(a+bi)(a-bi)

a^2-abi+abi-b^2i^2

a^2+0         -b^2(-1)

a^2+-b^2(-1)

a^2+b^2

See no need to use the whole foil method; the middle terms cancel.

So we are multiplying top and bottom of your fraction by (-3+4i):

\frac{-3+5i}{-3-4i} \cdot \frac{-3+4i}{-3+4i}=\frac{(-3+5i)(-3-4i)}{(-3-4i)(-3+4i)}

So you will have to use the complete foil method for the numerator. Let's do that:

(-3+5i)(-3+4i)

First: (-3)(-3)=9

Outer::  (-3)(4i)=-12i

Inner: (5i)(-3)=-15i

Last: (5i)(4i)=20i^2=20(-1)=-20

--------------------------------------------Combine like terms:

9-20-12i-15i

Simplify:

-11-27i

Now the bottom (-3-4i)(-3+4i):

F(OI)L (we are skipping OI)

First:-3(-3)=9

Last: -4i(4i)=-16i^2=-16(-1)=16

---------------------------------------------Combine like terms:

9+16=25

So our answer is \frac{-11-27i}{25}{/tex] unless you want to seprate the fraction too:[tex]\frac{-11}{25}+\frac{-27}{25}i

6 0
3 years ago
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