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erica [24]
3 years ago
9

Jeremy bought 5.6 pounds of chocolate. Jeremy's friend Susan bought twice as much chocolate as he did. How much more chocolate d

id Susan have than Jeremy?
Mathematics
1 answer:
Slav-nsk [51]3 years ago
6 0
Well if Jeremy bought 5.6 pounds then that means that Susan bought 11.2 pounds. So simply subtracting 5.6 from 11.2 will give you your answer which is 5.6. Personally I think they need to cut back on the chocolate that’s a bit much
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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.

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