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exis [7]
3 years ago
15

Let xy=1 and let dy/dt=5 Find dx/dt when x=5

Mathematics
1 answer:
madreJ [45]3 years ago
3 0
<span>So the first step is to take the derivative of xy=1. Don't forget the product chain rule!

</span>xy=1&#10;\\x=5&#10;\\y= \frac{1}{x}=\frac{1}{5}&#10;\\ &#10;\\\frac{dy}{dt}=5&#10;\\&#10;\\x\frac{dy}{dt}+y\frac{dx}{dt}=0&#10;\\&#10;\\5\times5+ \frac{1}{5} \frac{dx}{dt}=0&#10;\\&#10;\\\frac{1}{5} \frac{dx}{dt}=-25&#10;\\&#10;\\  \frac{dx}{dt}=-125&#10;<span>

</span>
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What is the missing number below in the solution to 968 divided by 10
Advocard [28]

Answer: 484/5

Step-by-step explanation:

For this word problem, it seems confusing with the wording but the equation would just be:

968÷10=x

solving for x, we get 484/5.

6 0
2 years ago
A discounted ticket for a football game costs $12.50 less than the original price.you pay $63 for a discounted ticket
Tanya [424]

Answer:

the original price is 75.50

Step-by-step explanation:

4 0
3 years ago
What is the length of side AD?. . . 6 cm. 8.5 cm. 10.4 cm. 12 cm
ludmilkaskok [199]
In triangle ABC, AC = 12/ (sin30) = 12 / (1/2) = 24
DC = 24-x
DB = DC tan 30 = (24-x)  tan30 <span>=(24−x)/</span><span>√3
</span>

In triangle ADB using Pythagorean Theorem<span><span>x2</span>+((24−x)/<span>√3</span><span>)2</span>=<span>12^2</span></span><span><span>x2</span>+(24−x<span>)^2</span>/3=<span>12^2</span></span><span>3<span>x2</span>+(24−x<span>)^2</span>=432</span><span>4<span>x2</span>−48x+576=432</span><span>4<span>x2</span>−48x+144=0</span><span><span><span>x2</span>−12x+36=0
x1 = x2 =6

AD = AC - DC = 24- (24-x) = 6</span></span>
5 0
3 years ago
Read 2 more answers
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
Sin (x/2)= √2- sin (x/2) solve for the exact solutions in the interval (0, 2π)
serg [7]

sin\left( \frac{x}{2} \right)=\sqrt{2}-sin\left( \frac{x}{2} \right)\implies sin\left( \frac{x}{2} \right)+sin\left( \frac{x}{2} \right)=\sqrt{2}\implies 2sin\left( \frac{x}{2} \right)=\sqrt{2} \\\\\\ sin\left( \cfrac{x}{2} \right)=\cfrac{\sqrt{2}}{2}\implies sin^{-1}\left[ sin\left( \cfrac{x}{2} \right) \right]=sin^{-1}\left( \cfrac{\sqrt{2}}{2} \right) \implies \cfrac{x}{2}= \begin{cases} \frac{\pi }{4}\\\\ \frac{3\pi }{4} \end{cases} \\\\[-0.35em] ~\dotfill

\cfrac{x}{2}=\cfrac{\pi }{4}\implies \boxed{x=\cfrac{\pi }{2}}~\hfill \cfrac{x}{2}=\cfrac{3\pi }{4}\implies \boxed{x=\cfrac{3\pi }{2}}

5 0
2 years ago
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