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Blizzard [7]
2 years ago
10

Find an equation for the line tangent to the curve y=x^3-4x+8 at the point (4,56)

Mathematics
1 answer:
Alexxandr [17]2 years ago
4 0
<span><span><span>y2</span>(<span>y2</span>−4)=<span>x2</span>(<span>x2</span>−5)</span><span><span>y2</span>(<span>y2</span>−4)=<span>x2</span>(<span>x2</span>−5)</span></span><span> Multiplying the polynomials gets us to </span><span><span><span>y4</span>−4<span>y2</span>=<span>x4</span>−5<span>x2</span></span><span><span>y4</span>−4<span>y2</span>=<span>x4</span>−5<span>x2</span></span></span><span>. Taking the derivative with respect to </span><span>xx</span><span> gets us: </span><span><span>4<span>y3</span><span>y′</span>−>!8y<span>y′</span>=4<span>x3</span>−10x</span><span>4<span>y3</span><span>y′</span>−>!8y<span>y′</span>=4<span>x3</span>−10x</span></span><span>. Factoring to get </span><span><span>y′</span><span>y′</span></span><span> by itself: </span><span><span><span>y′</span>(4<span>y3</span>−8y)=4<span>x3</span>−10)</span><span><span>y′</span>(4<span>y3</span>−8y)=4<span>x3</span>−10)</span></span><span>. Divide through to get </span><span><span>y′</span><span>y′</span></span><span> by itself: </span><span><span><span>y′</span>=<span><span>4<span>x3</span>−10x</span><span>4<span>y3</span>−8y</span></span></span><span><span>y′</span>=<span><span>4<span>x3</span>−10x</span><span>4<span>y3</span>−8y</span></span></span></span><span>. You could make your life a bit easier by factoring this into </span><span><span><span>y′</span>=<span><span>2x(2<span>x2</span>−5)</span><span>4y(<span>y2</span>−2)</span></span></span><span><span>y′</span>=<span><span>2x(2<span>x2</span>−5)</span><span>4y(<span>y2</span>−2)</span></span></span></span><span>. You could cancel out a factor of </span><span>22</span><span> to get </span><span><span><span>y′</span>=<span><span>x(2<span>x2</span>−5)</span><span>2y(<span>y2</span>−2)</span></span></span><span><span>y′</span>=<span><span>x(2<span>x2</span>−5)</span><span>2y(<span>y2</span>−2)</span></span></span></span><span>. To find the slope, plug in your points </span><span><span>x=0,y=−2</span><span>x=0,y=−2</span></span><span> into our equation for </span><span><span>y′</span><span>y′</span></span><span> to find the slope of the line. Note that the slope is </span><span>00</span><span>. To find the </span>equation<span> of the tangent line, use that value for </span><span>mm</span><span> you just found (</span><span><span>m=0</span><span>m=0</span></span><span>) and your given points into the point-slope formula and you find that the tangent line is </span><span><span>y=−2</span><span>y=−2</span></span><span>.

Thats what my Aunt said... Idk</span>
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Answer:

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Step-by-step explanation:

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3 years ago
Three girls cut out decimal models of hundredths. Amber doesn't shade any of her models. Grace qualifies three fifths of a model
Varvara68 [4.7K]

Answer:

Amber = 0

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Marsha =1.70

Step-by-step explanation:

Given

<em>Decimal Grid = Hundredths</em>

Amber = None

Grace = \frac{3}{5}th

Marsha = 1 + \frac{7}{10}th

To solve for each:

We have:

For Amber:

Amber = 0 * 100

Amber = 0

For Grace:

Grace = \frac{3}{5}th

Grace = \frac{3}{5} * 100

Grace = \frac{300}{5}

Grace = 60

This means that Grace shaded 60 of the hundredth models.

Decimal shaded is:

Grace = 60 * \frac{1}{100}

Grace =\frac{60}{100}

Grace = 0.60

For Marsha:

Marsha = 1 + \frac{7}{10}th

Marsha = \frac{10 + 7}{10}th

Marsha = \frac{17}{10}th

Marsha = 1.7th

Multiply this by 100, to get the number of shaded

Marsha = 1.7 * 100

Marsha = 170

This means that Marsha shaded 170 of the hundredth models.

Decimal shaded is:

Marsha = \frac{170}{100}

Marsha =1.70

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julsineya [31]

Answer:

7.5

Step-by-step explanation:

multiply like normal math . then add the decimal point after 1 number

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3 years ago
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Ksivusya [100]

Answer:

240 minutes

Step-by-step explanation:

If it's 40 minutes for just 5 papers, we need to figure how many minutes it would take to grade just 1. So you have to divide 40 by 5 which is 8. Then 8 time 30 which is 240.

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