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Ivanshal [37]
3 years ago
12

May I please get help in 7th grade math?

Mathematics
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

what

Step-by-step explanation:

what do you need help on?

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What is the domain and range of this function?
max2010maxim [7]

As for the domain, the only restriction comes from the logarithm. The argument of a logarith must be strictly positive, so we have

x-2>0 \iff x>2

As for the range, we have:

  • The range of \log(x) are all real numbers
  • If we change to \log(x-2) we're translating the function horizontally, so the range remains the same
  • If we change to \log(5(x-2() we're stretching the function horizontally, so the range doesn't change
  • If we change to \log(5(x-2))+1 we're translating the function 1 unit up, but the range is already all the real numbers, so it doesn't change.

5 0
3 years ago
30008876543*98765435678
Sladkaya [172]

I got math error ahhhaha

8 0
3 years ago
Read 2 more answers
I don't understand the concept of how to fine the answer ??
igor_vitrenko [27]
You need to find out how many pints are in 2,4,6,8 quarts. by the way 1 quart equals 2 pints. so to start you off. 2 quarts would be 4 pints and so on
4 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
4 years ago
Please help me. Given the explicit formula for an arithmetic sequence find the 52nd term.
ivann1987 [24]

Answer:

a = -41 +30n

52nd term is

a52 = -41 + 30(52)

a52 = -41 + 1560

a52 = 1519

3 0
3 years ago
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