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Llana [10]
3 years ago
5

Find the value of (–3⁄8) × (+8⁄15).

Mathematics
1 answer:
Otrada [13]3 years ago
8 0
<h2>-1/5</h2>

Step-by-step explanation:

\frac{-3}{8} × \frac{+8}{15}

= \frac{-1}{5}

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Solve the equation for x. Enter your answers as radical expressions. Then estimate to one decimal place, if necessary.
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Hope you won't be turned off by a correction, but we really need to use the symbol " ^ " to denote exponentiation.

Thus, we have x^2 = 100

Taking the square root of both sides, we get x = plus or minus 10. Verify, please, that both x = -10 and x = +10 satisfy the given equation.

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What is the probability of rolling 2 standard dice which sum to 11 ?
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1236

Step-by-step explanation:

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3 years ago
What is the vertex of the parabola with the equation y=1/5x^2 + 2x - 8
bixtya [17]

Answer:

(-5,-13)

Step-by-step explanation:

Find the vertex of the parabola y=1/5x^2+2x−8.  

In this equation a=15  and b = 2.  

x= -2/2(1/5) = -2/2/5 = -2/1 * 5/2 = -5

Substitute −5 into the equation y=1/5x^2+2x−8

y=1/5(−5)^2+2(−5)−8

y=5−10−8

y=−13

The vertex is (−5 , −13)

8 0
3 years ago
Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering
Svetllana [295]

Answer:

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Step-by-step explanation:

We are asked to find the tangent line approximation for f(x)=\sqrt{10+x} near x=0.

We will use linear approximation formula for a tangent line L(x) of a function f(x) at x=a to solve our given problem.

L(x)=f(a)+f'(a)(x-a)

Let us find value of function at x=0 as:

f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}

Now, we will find derivative of given function as:

f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}

f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)

f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1

f'(x)=\frac{1}{2\sqrt{10+x}}

Let us find derivative at x=0

f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}

Upon substituting our given values in linear approximation formula, we will get:

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)  

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Therefore, our required tangent line for approximation would be L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x.

8 0
3 years ago
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