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Blizzard [7]
3 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]3 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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9. Given the point (6,-8) values of the six trig function.
lozanna [386]

Answer:

Here's what I get.

Step-by-step explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10  

\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}

3 0
2 years ago
∫<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bxdx%7D%7B%28x%5E%7B2%7D%2B4%29%5E%7B3%7D%20%7D" id="TexFormula1" title="\frac{xdx}{
photoshop1234 [79]

Substitute u=x^2+4 and du=2x\,dx. Then the integral transforms to

\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \int \frac{du}{u^3}

Apply the power rule.

\displaystyle \int \frac{du}{u^3} = -\dfrac1{2u^2} + C

Now put the result back in terms of x.

\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \left(-\dfrac1{2u^2} + C\right) = -\dfrac1{4u^2} + C = \boxed{-\dfrac1{4(x^2+4)^2} + C}

3 0
2 years ago
The sale amount is 1,250.00, 5% sales tax.
Marina CMI [18]

Answer:

1250*.05=62.5. Then just add it to the amount for the total dollars shw needs to spend with tax

Step-by-step explanation:

7 0
3 years ago
1st question I want to solve
mr_godi [17]

Answer:

34.64

Step-by-step explanation:

take tan value and solve the triangle

7 0
2 years ago
What is 3 5/8 ÷14 please answer this
MissTica

Answer:

29/112

Step-by-step explanation:

3 5/8=29/8

(29/8)/14

(29/8)(1/14)=29/112

5 0
3 years ago
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