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Katarina [22]
2 years ago
9

Differentiate with respect to t

Mathematics
1 answer:
True [87]2 years ago
8 0

Answer:

\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0

Step-by-step explanation:

We want to differentiate the equation:

x^2-y^3+z^4=1

With respect to <em>t</em>, where <em>x, y, </em>and <em>z</em> are functions of <em>t. </em>

<em />

So:

\displaystyle \frac{d}{dt}\left[x^2-y^3+z^4\right]=\frac{d}{dt}\left[1\right]

Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:

\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0

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A triangle has angles measuring 15° and 45°. What is the measurement of the triangle’s third angle?
nexus9112 [7]
First you need to know that the internal angles of a triangle always equals 180°. Therefore:

15+45= 90°

Then since the angles should equal 180, you do:

180-90=90

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Stephen earned scores of 88 points, 94 points. 89 points, and 85 points on four tests.
Inessa05 [86]

answer

94

set up equations

average = total points / number of tests

total points = test1 + test2 + test3 + test4+ test5

total points = 88 + 94 +89 + 85 + test5

total points = 356 + test5

plug total points into first eqation and rearrange

average = total points / number of tests

average = (356 + test5)/number of tests

average * number of tests = 356 + test5

test5 = average * number of tests - 356

plug in values for average and number of tests

test5 = average * number of tests - 356

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3 0
3 years ago
Sketch the region enclosed by y=3x and y=x^2 . Then find the area of the region.
kolezko [41]

Answer:

A=4.5u^2

Step-by-step explanation:

The integral of a function gives you the area under the curve, the subtraction of one of the areas from the other will give you the area in between.

The limits of integration are the points where the curves intersect each other(take the curves has a system of equation and solve for x and y):

y=3x, y=x^2\\3x=x^2\\x^2-3x=0\\x(x-3)=0\\x_1=0\\x-3=0\\x_2=3

y_1=3x_1=3(0)=0\\y_2=3x_2=3(3)=9

The integral will be the subtraction of the curve y=3x and y=x^2  (In the graph you can see y=3x is the upper curve):

\int\limits^3_0 {3x-x^2} \, dx =\left\frac{3x^2}{2}-\frac{x^3}{3}\right|^3_0=\frac{3(3)^2}{2}-\frac{(3)^3}{3}-\left(\frac{3(0)^2}{2}-\frac{(0)^3}{3}\right)\\\int\limits^3_0 {3x-x^2} \, dx =\frac{27}{2}-\frac{27}{3} =13.5-9=4.5u^2

6 0
2 years ago
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