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Over [174]
3 years ago
6

What is the solution to the inequality 3t+>15

Mathematics
2 answers:
aev [14]3 years ago
7 0

Answer:

\large\boxed{t>5\to t\in(5,\ \infty)}

Step-by-step explanation:

3t>15\qquad\text{divide both sides by 3}\\\\t>5

umka2103 [35]3 years ago
5 0

Answer:

t > 5

Step-by-step explanation:

3t>15

Divide each side by 3

3t/3 > 15/3

t > 5

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Find the value of 1<br> and 1/10-1/5
Andreyy89

9514 1404 393

Answer:

  9/10

Step-by-step explanation:

(1 1/10) - 1/5 = (1 1/10) -2/10 = 1 + (1/10 -2/10) = 1 -1/10 = 9/10

__

1.1 -0.2 = 0.9 = 9/10

5 0
3 years ago
SHOW WORK I’ll mark you brainlist
Yuri [45]

Answer:

the answer is 0 for the reason being that it is a vertical slope.

change over y/change over x

1/0=0

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3 years ago
A bottle and a cap together cost $1.10. The bottle costs $1 more than the cap. How much does each cost?
Darya [45]

Answer:

The cap costs 10 cents and the botttle costs $1

Step-by-step explanation:

1.10-1=.10

3 0
3 years ago
A group of students went on a field trip to the zoo. 12 students went to the tiger
-BARSIC- [3]

Answer:

12:4, 12 to 4, 12/4

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tiger exhibit : gorilla exhibit

tiger exihbit = 12 students

gorilla exhibit = 4 students

8 0
4 years 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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