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koban [17]
3 years ago
14

How do I get a square root

Mathematics
1 answer:
Maslowich3 years ago
4 0

Answer:

find the closest perfect square root to the number you are trying to solve.

to get a square root, you need to find the number that multiplied by itself will give you the number.

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\sqrt{x+7}     x = 9

Plug in 9 for "x" in the equation

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3 years ago
What is the equivalent fraction for 5/40
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4 years ago
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What is the exact value of cos 105 degrees?
Fittoniya [83]
Your answer is C c:

First, split the angle into two angles where the values of the six trigonometric functions are known. In this case, 105, can be split into 45+60 (cos (45+60)) 

Use the sum formula for cosine to simplify the expression. The formula states that cos (A+B) = - (cos (A) cos (B) + sin (A) sin (B)).

The exact value of cos(60) is 1/2. So multiply 1/2 to cos (45) - sin (60) times sin (45).

The exact value of cos (45) is the square root of 2 over 2 

(1/2) ⋅ (<span>√2/</span>2) - sin (60) <span>⋅ sin (45)
</span>
Value of sin (60) is <span>√3/2

</span>(1/2) ⋅ (√2/2) - √3/2 ⋅ sin (45)

exact value of sin (45) is √2/2

(1/2) ⋅ (√2/2) - √3/2 ⋅ √2/2 

all that equals √2/4 - √6/4


3 0
4 years ago
Pls help only if yk it ill give brainliest to the correct answer !
IgorC [24]

Answer:

x=-0.3

x=-11.7

Step-by-step explanation:

4x^2+24x+13=2x^2+6\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\4x^2+24x+13-6=2x^2+6-6\\Simplify\\4x^2+24x+7=2x^2\\\mathrm{Subtract\:}2x^2\mathrm{\:from\:both\:sides}\\4x^2+24x+7-2x^2=2x^2-2x^2\\Simplify\\2x^2+24x+7=0\\\mathrm{Solve\:with\:the\:quadratic\:formula}\\Quadratic\:Equation\:Formula\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=24,\:c=7:\quad x_{1,\:2}=\frac{-24\pm \sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}\\x=\frac{-24+\sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}:\quad \frac{-12+\sqrt{130}}{2}\\x=\frac{-24-\sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}:\quad -\frac{12+\sqrt{130}}{2}\\The\:solutions\:to\:the\:quadratic\:equation\:are:\\x=\frac{-12+\sqrt{130}}{2},\:x=-\frac{12+\sqrt{130}}{2}\\\\quad x=-0.29912\ ,\:x=-11.70087

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3 years ago
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