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8090 [49]
2 years ago
8

Find the Two Values that x can have if X^2 -10=26​

Mathematics
2 answers:
Dmitry [639]2 years ago
6 0

Answer:

+6 and -6.

Step-by-step explanation:

x^2 - 10 = 26​

x^2  = 26  + 10 = 36

x = +/- 6.

Zolol [24]2 years ago
3 0

Answer:

x1= -6 x2=6

Step-by-step explanation:

x^2-10 = 26

x^2 = 26 +10

x^2 = 36

x = sqrt(36)

x = -6 or x = 6 because both 6 and -6 squared are equal to 36

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My Math Integers. I'm stuck on this problem.. Someone please help? (-4)(-9) =
tatuchka [14]

Answer:

It is 36

Step-by-step explanation:

minus × minus equals + so its 36

5 0
2 years ago
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Find an irrational number between -10 and -10.1.
Pavlova-9 [17]
   
Between two rational numbers there are an infinite number of irrational numbers.

Example:


- 5 \sqrt{3} -\sqrt{2}= 5 \times 1.7320508 - 1.41421356 \approx -10.0744676\\\\
-10.1 \ \textless \  -10,0744676 \ \textless \  -10



6 0
2 years ago
What is the square root of 1,024
Shkiper50 [21]
1024 can be factored into 
<span>2(512) </span>
<span>=2^2(256) </span>
<span>=2^3(128) </span>
<span>=2^4(64) </span>
<span>=2^5(32) </span>
<span>=2^6(16) </span>
<span>=2^10 </span>

<span>Therefore, </span>
<span>√1024 = √(2^10) = 2^5 = 32</span>
4 0
2 years ago
HELLPPPPPP!!!
lilavasa [31]

Answer:

B - 181/500

Step-by-step explanation:

To convert percentages to fractions you divide by 100. That would make this 36.2/100. But we dont like fractions with decimal points in so you would multiply the whole thing by 5 to get rid of the .2 this makes the answer 181/500

3 0
2 years ago
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Determine the zeroes of the polynomial
Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
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