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____ [38]
2 years ago
5

Find the value of x in the triangle shown below.

Mathematics
1 answer:
S_A_V [24]2 years ago
6 0

triangle is a good shape... but

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Which choice is equivalent to the quotient shown here when x is greater than or equal to 0?
Dima020 [189]

Answer: OPTION C

Step-by-step explanation:

Remember that:

\sqrt[n]{a^n}=a

And the Product of powers property establishes that:

a^m*a^n=a^{(mn)}

Rewrite the expression:

\frac{\sqrt{18x} }{\sqrt{32} }

Descompose 18 and 32 into their prime factors:

18=2*3*3=2*3^2\\32=2*2*2*2*2=2^5=2^4*2

Substitute into the expression, then:

\frac{\sqrt{(2*3^2)x} }{\sqrt{2^4*2} }

Finally,simplifying, you get:

\frac{3\sqrt{(2)x} }{2^2\sqrt{2} }=\frac{3\sqrt{2x}}{4\sqrt{2}}=\frac{(3)(\sqrt{x})(\sqrt{2})}{(4)(\sqrt{2})}= \frac{3\sqrt{x}}{4}

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3 years ago
Write a equation of a hyperbola given the foci and the asymptotes
professor190 [17]

Solution:

The standard equation of a hyperbola is expressed as

\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\text{ \lparen parallel to the x-axis\rparen} \\ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ \lparen parallel to the y-axis\rparen} \end{gathered}

Given that the hyperbola has its foci at (0,-15) and (0, 15), this implies that the hyperbola is parallel to the y-axis.

Thus, the equation will be expressed in the form:

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ ----equation 1}

The asymptote of n hyperbola is expressed as

y=\pm\frac{a}{b}(x-h)+k

Given that the asymptotes are

y=\frac{3}{4}x\text{ and y=-}\frac{3}{4}x

This implies that

a=3,\text{ and b=4}

To evaluate the value of h and k,

undefined

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Answer:4566878888

Step-by-step explanation:

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Four runners start at the same point: Lin, Elena, Diego, Andre. For each runner write a multiplication equation that describes t
jeka94

Answer:

I honestly have no clue. Sorry I cant help you

Step-by-step explanation:

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Whats greater .007 or .07
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.07 is greater than .007
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