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ExtremeBDS [4]
2 years ago
8

2. Solve for y. A 10.5 B 9 C 13.5 D 12

Mathematics
1 answer:
Vera_Pavlovna [14]2 years ago
5 0
It’s A
Because if you subtract 12 from 96 you’ll get 8y=84 then divide both sides by 8 you get 21/2 simplify and get 10.5 or 10 1/2
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PLz i need help will mark brianly
gulaghasi [49]
Since both the expressions in x are equal to y then they are equal to each other, so

-2x + 3 = 13x - 4

-15x + 3 = -4

-15x = -7 

x = 7/15

and y = -2(7/15) + 3 =  31/15  = 2 1/15
3 0
3 years ago
Read 2 more answers
If you earn $15 an hour and you work 35 hours a week, how much do you earn a month?
Ber [7]
2100$ you earn a month because you multiply 15 and 35 and u get 525 then multiply that by 4 since that’s how many weeks are in a month and u get 2100
7 0
3 years ago
Solve for x. (30/x) = 4/6
andre [41]
X=30×6/4
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4 0
3 years ago
NEED HELP
blagie [28]

Answer:

Range: [-7, 8]

General Formulas and Concepts:

<u>Algebra I</u>

  • Reading a coordinate plane
  • Range is the set of y-values that are outputted by function f(x)
  • Interval Notation: [Brackets] denote inclusion, (Parenthesis) denote exclusion

Step-by-step explanation:

According to the graph, our y-values span from -7 to 8. Since both are closed dot, they are included in the range:

Range: [-7, 8]

3 0
3 years ago
Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an
Leya [2.2K]
Parameterize the lateral face T_1 of the cylinder by

\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v

where 0\le u\le2\pi and 0\le v\le3, and parameterize the disks T_2,T_3 as

\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)
\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)

where 0\le r\le2 and 0\le\theta\le2\pi.

The integral along the surface of the cylinder (with outward/positive orientation) is then

\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS
=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr
=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr
=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr
=136\pi
7 0
3 years ago
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