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wlad13 [49]
3 years ago
5

Evaluate the integral f*dr where f(x,y)=<-y,x^2> and c consists of te arc of the parabola y=2x^2 from (-1,2) to (1,2)

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
3 0
y=2x^2\implies\mathbf r(x)=\langle x,2x^2\rangle

\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_{t=-1}^{t=1}\langle-2x^2,x^2\rangle\cdot\langle1,4x\rangle\,\mathrm dx
=\displaystyle\int_{-1}^1(8x^3-2x^2)\,\mathrm dx
=-\dfrac43
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Which of the following are solutions of 2x &lt; 26?
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Step-by-step explanation:

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Read 2 more answers
Drag each tile to the correct box.
Natasha_Volkova [10]

Answer:

1) Function h

interval [3, 5]

rate of change 6

2) Function f

interval [3, 6]

rate of change 8.33

3) Function g

interval [2, 3]

rate of change 9.6

Step-by-step explanation:

we know that

To find the average rate of change, we divide the change in the output value by the change in the input value

the average rate of change is equal to

\frac{f(b)-f(a)}{b-a}

step 1

Find the average rate of change of function h(x) over interval [3,5]

Looking at the third picture (table)

f(a)=h(3)=4  

f(b)=h(5)=16

a=3

b=5

Substitute

\frac{16-4}{5-3}=6

step 2

Find the average rate of change of function f(x) over interval [3,6]

Looking at the graph

f(a)=f(3)=10  

f(b)=f(6)=35

a=3

b=6

Substitute

\frac{35-10}{6-3}=8.33

step 3

Find the average rate of change of function g(x) over interval [2,3]

we have

g(x)=\frac{1}{5}(4)^x

f(a)=g(2)=\frac{1}{5}(4)^2=\frac{16}{5}  

f(b)=g(3)=\frac{1}{5}(4)^3=\frac{64}{5}

a=2

b=3

Substitute

\frac{\frac{64}{5}-\frac{16}{5}}{3-2}=9.6

therefore

In order from least to greatest according to their average rates of change over those intervals

1) Function h

interval [3, 5]

rate of change 6

2) Function f

interval [3, 6]

rate of change 8.33

3) Function g

interval [2, 3]

rate of change 9.6

7 0
3 years ago
Need help on this math study guide
photoshop1234 [79]

8.

2 terms on left side, 3 on right side

9.

All variables count as coefficients,  so they’d be 8m, k, and -16k

10.

The constants are just numbers on their own, or without a variable next to it. Here it’d be 10 and -4


6 0
3 years ago
write an equation of the line passing through point p that is perpendicular to the given line p(6,10),y=-3x+13
patriot [66]

Answer:

y = \frac{1}{3} x + 8

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

y = - 3x + 13 is in this form

with m = - 3

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{-3} = \frac{1}{3}

y = \frac{1}{3} x + c ← is the partial equation

To find c substitute (6, 10) into the partial equation

10 = 2 + c ⇒ c = 10 - 2 = 8

y = \frac{1}{3} x + 8 ← equation of line


6 0
3 years ago
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