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3 years ago

Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin

1 answer:

3 0

Y = xe^x
dy/dx(e^x x)=>use the product rule, d/dx(u v) = v*(du)/(dx)+u*(dv)/(dx), where u = e^x and v = x:
= e^x (d/dx(x))+x (d/dx(e^x))
y' = e^x x+ e^x
y'(0) = 1 => slope of the tangent
slope of the normal = -1
y - 0 = -1(x - 0)
y = -x => normal at origin

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Over the past seven days, Mrs. Cho found that the temperature outside had dropped a total of 35

s344n2d4d5 [400]

Answer:

Average change in temperature each day = 5 degrees/day

Step-by-step explanation:

$\frac{35}{7} = 5$

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3 years ago

Joanna purchased four boxes of cookies for a party. Each box cost $3.28, including tax. PLEASE HELP!!! 100 POINTS!!! Which expre

Olegator [25]

Answer:

4 / ($3.28) is the correct answer

Step-by-step explanation:

As cost of one box is $3.28

the cost of four boxes will be 4 divided by the cost of one box....so that why the 4/($3.28) is correct option

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3 years ago

Ariana is shipping 8 bags of granola to a customer. Each bag weighs 22 ounces and the maximum weight she can ship in one box is

Anuta_ua [19.1K]

Ariana cannot ship all the bags in one box.

Ariana will be sending 8 bags which will weigh a total of:

= 22 ounces x 8

= 176 ounces

You need to convert this to pounds to see if it will fit in the box:

<em>1 pound = 16 ounces </em>

176 ounces in pounds is therefore:

= 176 / 16

= 11 pounds

The box can only take 10 pounds and 5 ounces which is less than the 11 pounds the bags weigh.

Ariana cannot ship all the bags in one box.

<em>Find out more at brainly.com/question/19990206.</em>

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3 years ago

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

But, angle $t$ is in 3rd quadrant, so value of

$\bold{cot(t) =\dfrac{12}{5}}$

4 0

3 years ago

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lozanna [386]

tell the height, width and depth then we will help

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3 years ago