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2 years ago

Help please ! i need this asap

Mathematics

2 answers:

Novay_Z [31]2 years ago

7 0

The person above is correct it’s 125

Tanzania [10]2 years ago

6 0

Answer:

x = 65° + 60° = 125°

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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

2 years ago

Find the solution of this system of equations?<br><br> -8x-6y=60<br><br> 5x+6y=-69

zvonat [6]

First subtract the x variable on both sides so on the first equation youll have -6y=8x+60 then divide 6 on all variables which means youll have y=8/6x+10 and in the second equation you do the same thing and youll have y=-5/6x-11.5

7 0

2 years ago

Read 2 more answers

How do you calculate the volume for this shape

Bingel [31]

Answer:

The formula A=12bh is used to find the area of the top and bases triangular faces, A= area, b= base, and h= height. The formula A=lw is used to find the area of the three retangular side faces, A= area, l= lenght, and w= width

Step-by-step explanation:

3 0

2 years ago

Lezlie hiked 3/8 of a mile on Monday. On Wednesday she hiked 3/6 of a mile. She hiked 3/4 of a mile on Friday. Use benchmark num

Gnoma [55]

She walked the most on Friday because you have to find the LCD. The LCD is 24.
For example, to change the denominator of 4 to 24, you must divide 24 divided by 4 which equals 6. Since your answer was 6, you have to multiply both numbers, which are 3 and 4, by 6.
So, 4 times 6 is 24. This is the denominator and 4 times 3 is 18. This is the numerator.
If you do the same with all of the fractions then you will have the fractions of, 18 over 24, 12 over 24, and 9 over 24.
Look at all the numerators (the number on top of the fraction) and decide which one is the biggest.
18 is the biggest and since the fraction 18 over 24 was originally 3 over 4, this means that she walked the most on Friday.
Hoped this helps!!

3 0

3 years ago

The value 5 is an upper bound for the zeros of the function shown below.

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$