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4 years ago

24 POINTS!!!!!!!!!!

Mathematics

2 answers:

maw [93]4 years ago

5 0

If the signs are alike, and ur adding/subtracting......u add them and keep the same sign

finlep [7]4 years ago

3 0

Adding or subtracting

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What is the value of 9^-1/2

qaws [65]

Answer:

81-1/2

80/2

40 and right than say

7 0

3 years ago

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The value of polynomial 5x-4x^2+3, when x=-1<br>

Alchen [17]

Answer:

-6

Step-by-step explanation:

$\text{Given that,}\\\\5x-4x^2 +3 \\\\\text{When}~ x = -1,\\\\~~~5(-1) -4(-1)^2 +3 \\\\=-5-4(1)+3\\\\=-5-4+3\\\\=-5-1\\\\=-6$

8 0

2 years ago

The sum of two numbers is at most 12, and the sum of 3 times the first number and 8 times the second number is at least 48 selec

Sever21 [200]

Answer:

(4,8)

Step-by-step explanation:

4 + 8 = 12

3(4) + 8(8)

= 12 + 64

= 76

6 0

3 years ago

If 9% of a number equals 10, find 90% of that number.

rodikova [14]

The answer is 100 % because it’s the same but with two zeros

5 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago