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PolarNik [594]
3 years ago
12

For the following​ equation:

Mathematics
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

a. 0

b. x = 1.25

Step-by-step explanation:

The given equation is:

\frac{4}{x}=\frac{1}{4x}+3

a. The denominators are x and 4x. The values that make a denominator zero are:

x=0\\4x=0\\x=0

b. Solving the equation:

\frac{4}{x}=\frac{1}{4x}+3\\\frac{4x}{x}=\frac{x}{4x}+3x\\  3x=4-\frac{1}{4}\\x=\frac{16-1}{4*3} \\ x=1.25

The solution is x = 1.25

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Write the number in two other forms 6.7
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Word form - six.seven Expand form - 6 + 0.7
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Elena said the equation 9x+15=3x+15 has no solutions because 9x is greater than 3x. Do you agree with Elena? Explaining your rea
erastova [34]

Answer:

it does have a solution x=0

Step-by-step explanation:

9x+15=3x+15

6x+15=15

6x=0

x=0

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Ryan said that 212.314 73.243 285.557. Is Ryan's answer reasonable?
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On a map, 1/4 inch between locations represents an actual distance of 2 miles between the locations. What is the actual distance
Fittoniya [83]

Answer:

12/17 miles

Step-by-step explanation:

Make the denominators equal -> 1/4 = 17/68, 3/34 = 6/68

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The ratio between 17/68 and 6/68 is 17/68 : 6/68 = 17/6
So, the ratio between the distance should also be 17/6

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Hope it helps :)

3 0
2 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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