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ziro4ka [17]
3 years ago
7

If (x + 1/x)^2 =9, then (x-1/x)^2= A) 3 B) 5 C) 7 D 9

Mathematics
1 answer:
nataly862011 [7]3 years ago
4 0

Answer:

c se não for seta mi disputa amiga

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In a yoga class, 75% of the students are women. There are 24 women in the class. How many students are in the class?
almond37 [142]

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32

Step-by-step explanation:

8 0
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What can you triple and then subtract by 7 to get -1
iVinArrow [24]

Answer:

2

Step-by-step explanation:

<em>let </em><em>the </em><em>number </em><em>be </em><em>x</em>

<em>The</em><em>n</em><em> </em><em>you </em><em>triple </em><em>the </em><em>x </em><em>and </em><em>minus </em><em>7 </em><em> </em><em>from </em><em>it </em><em>and </em><em>you </em><em>equate </em><em>it </em><em>to </em><em>-</em><em>1</em>

7 0
3 years ago
Solve<br> y=-x2+2x+10<br> y=x+4
Marizza181 [45]
Y=10
explanation:
-4•2= -8
2•4=8
-8+8=0
y=10
4 0
3 years ago
[<br> 1<br> What is the distance between the points (-4, -8) and (10,-8)?<br> Need help
Vedmedyk [2.9K]

Answer:

2

Step-by-step explanation:

√(-8-8)^2 +(10+4)^2=

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3 0
3 years ago
Which equation results from taking the square root of both sides of (x 9)2 = 25? x 3 = ±5 x 3 = ±25 x 9 = ±5 x 9 = ±25.
kow [346]

The equation which is results from taking the square root of both sides of the provided equation is,

(x+9)=\pm5\\

<h3>What is the square root?</h3>

A square root of a number is the value which is when multiplicand by itself gives the same value as the number posses.

Let a number is <em>a. </em>Then this number in the form of square root can be written as,

a=\sqrt{a}\times \sqrt{a}

The given algebraic equation in the problem is,

(x+ 9)^2 = 25

Take the square root, in both sides,

\sqrt{(x+ 9)^2} = \sqrt{25}\\\sqrt{(x+ 9)^2} = \sqrt{5^2}

Cancel out the square root with the square of the number as,

(x+9)=\pm5\\

Hence, the equation which is results from taking the square root of both sides of the provided equation is,

(x+9)=\pm5\\

Learn more about the square root of number here;

brainly.com/question/664132

7 0
3 years ago
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