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nignag [31]
2 years ago
5

Find the length of side X

Mathematics
1 answer:
NISA [10]2 years ago
7 0

Answer:

C

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

The square on the hypotenuse is equal to the sum of the squares on the 2 other sides, that is

x² + 6² = 17²

x² + 36 = 289 ( subtract 36 from both sides )

x² = 253 ( take the square root of both sides )

x = \sqrt{253} → C

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Answer:

c

Step-by-step explanation:

just cause

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3 years ago
Find the equation of the line that is perpendicular to 3x + 2y = -2 and goes through the point (6,2)
notka56 [123]
Perpendicular lines have slopes that are inverse reciprocals of each other. That would give you the new slope in your equation. After that plug (6,2) into point slope form and solve to get your final equation.

4 0
2 years ago
Is the expression 5 (3m+2p) -4m equivalent to the expression 13p+11m-3p?
Aleksandr [31]
 5 (3m+2p) -4m= 15m +10p -4m
=11m +10p
--------------------------------------------------------
13p+11m-3p=10p -11m
-----------------------------------------------------------
Yes, 5 (3m+2p) -4m is  equivalent to the expression 13p+11m-3p as i worked out for you.. Hope this helps you!!! =')

7 0
3 years ago
Read 2 more answers
What are the intercepts of the equation 18x-9y+3z=18
IceJOKER [234]
The answer is:  [D]:  " <span>(1, 0, 0), (0, –2, 0), (0, 0, 6) " .
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6 0
2 years ago
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Solve the rational function of (check image) and check for extraneous solutions
Oxana [17]

Answer:

Option A is correct.

i.e. x = 1, x = 0 is an extraneous solution.

Step-by-step explanation:

Given the expression

\frac{5}{x}=\frac{4x+1}{x^2}

Solving the rational function

\frac{5}{x}=\frac{4x+1}{x^2}

Apply fraction across multiply: if \frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5x^2=x\left(4x+1\right)

Subtract x(4x+1) from both sides

5x^2-x\left(4x+1\right)=x\left(4x+1\right)-x\left(4x+1\right)

Simplify

5x^2-x\left(4x+1\right)=0

5x² - 4x² - x = 0

x² - x = 0

Factor x² - x = x(x-1)

so

x(x-1) = 0

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

x=0\quad \mathrm{or}\quad \:x-1=0

Thus, the solution to the equation is:

x=0,\:x=1

But, it is clear that if we substitute x = 0, the equation becomes undefined because we can not have the denominator to be 0.

In other words, the equation is undefined for x = 0

Thus, x = 0 is an extraneous solutions.

Therefore, option A is correct.

i.e. x = 1, x = 0 is an extraneous solution.

5 0
3 years ago
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