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labwork [276]
2 years ago
6

Plz help me with this

Mathematics
1 answer:
Delvig [45]2 years ago
4 0

Answer:  C) log₄ 84

<u>Step-by-step explanation:</u>

The rule for condensing (combining) log expressions that are added is MULTIPLICATION

log₄12 + log₄7 = log₄(12 × 7) = log₄ 84

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The table below shows the different combinations of honey and oats that can be used in making granola.
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Answer:

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Which of the following is equivalent to (a + b/2) ^2?
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a^2 + 4a/b + 4/b^2

Step-by-step explanation:

Which of the following is equivalent to (a + b/2) ^2?

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Simplify √ 96/8 a.√2/4 b.1/2√3 c.4√3/√2 d. 2√3
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Answer:

Just substitute each number into the expression:

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Step-by-step explanation:

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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
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