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Annette [7]
2 years ago
8

20% of what number is 47? Need help

Mathematics
1 answer:
ser-zykov [4K]2 years ago
8 0
<span>47 : 20% = 47 / (20 : 100) = (100 * 47) / 20 = 235 If the case, values are rounded to ... Rewrite: 20% of the students means 7 students => 20% * Total number of ...</span><span>
</span>
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Solve a<br> a-20=b<br> a=<br><br><br><br> plzz help
saveliy_v [14]
In order to have the variable by itself you have to add 20 to both sides. So your answer would be a=b+20. I hope this helps love! :)
7 0
3 years ago
How would you do 5^-1 (mod 26) ?
Yuri [45]

What we’re looking for is the value of k that will satisfy the equation.

The solution would be like this for this specific problem:

5k =1 (mod 26)

Check how 5*5 = 25 = -1 (mod 26)

In this way, 5*(-5) = -5*5 = -25 = -(-1) = 1 (mod 26)

Thus, k = -5 = 21 (mod 26)

And 5^-1 = 21 (mod 26)

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

3 0
3 years ago
Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(
BlackZzzverrR [31]
We can parameterize this part of a cone by

\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle

with 0\le u\le a and 0\le v\le2\pi. Then

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv

The area of this surface (call it \mathcal S) is then

\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi
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2 years ago
At Winfield University, the student population is growing quickly. The university has two existing square student parking lots o
Mashutka [201]
The answer is D because it is
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2 years ago
I really need these answered!
Akimi4 [234]

Answer:

Step-by-step explanation:

#1) \arcsin \left(0.64958\right)=0.70703\dots \quad \begin{pmatrix}\mathrm{Degrees:}&40.51^{\circ \:}\end{pmatrix}

∡C =62.49

\frac{\left(\sin \left(27^{\circ \:}\right)\right)}{3}=\frac{\sin \left(54^{\circ \:}\right)}{x}\quad :\quad x=\frac{3\left(\sqrt{5}+1\right)}{\sin \left(27^{\circ \:}\right)\cdot \:4}\quad \left(\mathrm{Decimal}:\quad x=5.34603\dots \right)

6 0
2 years ago
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