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WARRIOR [948]
3 years ago
8

Factor the polynomial expression x4 + 18x2 + 81

Mathematics
1 answer:
Akimi4 [234]3 years ago
4 0

Answer: (x^2+9)^2

Step-by-step explanation:

Rewrite it in the form a^2+2ab+b^2 , where a=x^2 and b=9.

(x^2)^2+2(x^2)(9)+9^2

Use Square of Sum: (a+b)^2= a^2+2ab+b^2

(x^2+9)^2

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Find the m<br> A)50 degrees <br> B)60 degrees <br> C)70 degrees <br> D)80degrees
lara31 [8.8K]

Answer:

70

Step-by-step explanation:

65 and 45 make up 110, and triangles are composed up 180°. The missing angle thatll make it add up to 180 is 70

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3 years ago
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Brian flips a fair coin 3 times. What is the probability of getting exactly 2 tails?
sammy [17]

Answer:

3/8

Step-by-step explanation:

The possible outcomes

TTT TTH THT THH  HHH HHT HTH HTT  = 8 outcomes

There are TTH THT HTT  3 with exactly two tails

P ( exactly 2 tails) = 3/8

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Daisy and Sam scored baskets at the basketball game in the ratio of 3:2. Daisy scored 6 more baskets than Sam. How many baskets
Scorpion4ik [409]
Daisy scored 15 baskets
5 0
2 years ago
2. Mulligan Mall has a square-shaped ice-skating rink as shown below. In the middle of the ice-skating rink there is a snack bar
gladu [14]

Answer:

2167 meters squared

Step-by-step explanation:

First find the area of the square ice skating rink: 47^2 = 2209.

Then, find the area of the triangle without ice by using 1/2bh: (12/2)(7)=42.

Simply subtract the triangle's area from the squares for the final answer of 2167 meters squared.

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3 years ago
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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
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