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3 years ago

Solve for 5x+20=x-10 I’ll give brain if anyone can solve it please

Mathematics

1 answer:

Inessa05 [86]3 years ago

4 0

Answer: x = -15/2

Step-by-step explanation:

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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

3 years ago

What is the measure of arc BC?

Artist 52 [7]

9514 1404 393

Answer:

100°

Step-by-step explanation:

Arc BC is twice the measure of inscribed angle BEC, so is ...

arc BC = 2×50°

arc BC = 100°

5 0

3 years ago

Read 2 more answers

Need help pls .......

sukhopar [10]

Answer:

It should be the last one because rigid transformations doesn't change in size just where it is

7 0

3 years ago

Need help asap thanks

Ymorist [56]

Answer:

Step-by-step explanation:

for a right angled triangle, the sum of the square of the two sides should be equal to the square of the hypotenuse. If this rule does not hold, the dimensions do not represent the side of a right angled triangle.

The Pythagoras theorem : a² + b² = c²

where a = length

b = base

c = hypotenuse

4² + 10² ≠ 16²

6² + 12² = 13²

7² + 24² = 25²

8² + 15² = 17²

8 0

3 years ago

HELP ASAP PLEASEEEEEEEEEE

Andrew [12]

Given the recursive formula, each terms is five time the previous one.

This means that:

$f(6)$ is 5 times $f(5)$ , which means $f(6)=5f(5)$
in turn, $f(5)$ is 5 times $f(4)$ , so we have $f(5)=5f(4)$ . This means that $f(6)=5f(5)=5\cdot 5f(4) = 25f(4)$
Finally, $f(4)=5f(3)$ So, substituting this back gives

$f(6)=25f(4)=25\cdot 5f(3)=125f(3)$

In general, since you have $f(n)=5f(n-1)$ , each time you compute a new term you multiply by a factor of 5, so if $m=n+k$ , you have

$f(m)=5^kf(n)$

4 0

3 years ago