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3 years ago

What would be the slope-intercept form of the equation of the line through (0, 6) and (4, 4)?

Mathematics

1 answer:

Stolb23 [73]3 years ago

7 0

Answer:

i only know this

The equation of the line of slope 4 is y=4x+b. The line goes through (-1,-6) If you step 1 unit positive in the x direction you reach the y-axis, (x=0). In order to remain on the line you need to go up 4 units, that is from -6 to -2. y=4x-2.

Step-by-step explanation:

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A square room has a floor area of 81 square meters. The height of the room is 5 meters. What is the area of one wall

VMariaS [17]

Area of a square is side length ^2

To find the length of the side of a square from the area, that’s the square root of the area:

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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:

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

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

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

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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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3 years ago

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Answer:

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