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Vikki [24]
3 years ago
5

PLS HELP ILL GIVE BRAINLIEST

Mathematics
1 answer:
Blizzard [7]3 years ago
8 0

Hi there.

The answer is f(0) = 3.

Explanation:

Given 2 functions with given domains.

f(x) = 5x - 1

where x < -2

f(x) = x + 3

where x >= -2

Evaluate the value of f(x) when x = 0. Therefore, f(x)= 5x-1 cannot be used as the given x = 0 is greater than its domain.

Thus, our main function is f(x) = x+3

Evaluate the value by substituting x = 0 in the function.

f(0) = 0 + 3 \\ f(0) = 3

Thus, the answer is f(0) = 3

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Rewrite the expression 5(6 + 9) as the sum of 30 and another whole number.
KATRIN_1 [288]

Answer:

30 + 45

Step-by-step explanation:

5(6 + 9) = 5*6 + 5*9 = 30 + 45

5 0
3 years ago
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Solve for x.<br> x + 8 = 12
VashaNatasha [74]

Answer:

Step-by-step explanation:

move constant to the right and change the sign

X=12-8

X=4

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3 years ago
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Solve for x 12x/5 a.x=-71/2 b. x=71/2 c. x= 43 1/5
babunello [35]

Answer:

x  = - 12 6/7

Step-by-step explanation:

x - 12x/5= 18

Get a common denominator on the left

5/5 x -12/5x = 18

Subtract

-7/5 x = 18

Multiply each side by -5/7 to isolate x

-5/7 *-7/5 x = -5/7 *18

x = -5*18/7

x = -90/7

Changing this to a mixed number

7 goes into 90   1 2 times with 6 left over

x  = - 12 6/7


8 0
3 years ago
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Suppose that the cost of a rental car is $65 for a week plus 0.15 per mile driven.
IceJOKER [234]

You have the following equation for the cost of renting a car for x number of miles driven:

y = 0.15x + 65

If you travel 70 miles, then, x = 70 and you obtain for the cost y:

y = 0.15(70) + 65

y = 10.5 + 65

y = 75.5

Hence, the cost of renting a car to drive 70 miles is $75.5

3 0
1 year ago
Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the
ludmilkaskok [199]
By Stokes' theorem,

\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S

where \mathcal C is the circular boundary of the hemisphere \mathcal M in the y-z plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle

where 0\le t\le2\pi. Then the line integral is

\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt
=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi

We can check this result by evaluating the equivalent surface integral. We have

\nabla\times\mathbf f=\langle1,0,0\rangle

and we can parameterize \mathcal M by

\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle

so that

\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv

where 0\le v\le\dfrac\pi2 and 0\le u\le2\pi. Then,

\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi

as expected.
7 0
3 years ago
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