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Ira Lisetskai [31]
3 years ago
12

HELP ASAP WILL GIVE BRAINLY

Mathematics
1 answer:
Mila [183]3 years ago
8 0

Answer:

1 and 2 are not polyhedrons, as a circle has infinite sides

1. cone, 1 vertex

2. sphere, nothing

3. pentagonal prism?

wait but can't every face be a base???

anyway 7 faces, 10 vertices

4. triangular prism

5 faces, 6 vertices

Step-by-step explanation:

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What is the slope and and y-int of y≤3
Neporo4naja [7]

Answer:

0

I hope this is the correct answer

7 0
2 years ago
Read 2 more answers
Use distributive law evaluate
Cerrena [4.2K]

Answer:

The distributive property allows you multiply a sum in parenthesis by multiplying each addend separately, then add the products.

Step-by-step explanation:

How to use the distributive law example.

2(x+4) = 16

To use the distributive law in this example multiply 3 by all terms in the parenthesis. Multiply 2 and x, then 2 and 4 to open the parenthesis.

2x+8=16

That is how you use the distributive law.

To continue solving, subtract 8 from both sides.

2x+8-8=16-8

2x=8

Divide 2 from both sides.

2x/2=8/2

x=4

Hope this helps!

If not, I am sorry.

6 0
2 years ago
What is the equation for the graph shown below?
ValentinkaMS [17]

Answer: D

Explanation:

Since we don't know the y-value of the vertex, let's do this an easy way: plugging in.

Let's use the y intercept since that would be the easiest. Since x=0, the terms with x cancel, and you will get a leading result of -5

A) results in 2, so eliminate it

B) results in -2, so eliminate it

C) results in 5, so eliminate it

D) results in -5, so keep it

Since D was the only one that worked, that is our correct answer.

4 0
2 years ago
Can anyone help me integrate :
worty [1.4K]
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
3 0
3 years ago
Which number is a multiple of 7? A 27 B 48 C 56 ET D 74​
saul85 [17]

Answer:

C. 56

Step-by-step explanation:

7*8=56

7 0
2 years ago
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