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3 years ago

6x+4=45 find the value of x

Mathematics

1 answer:

RSB [31]3 years ago

8 0

Answer:

x = 41/6 or 6.8333333

Step-by-step explanation:

When solving for x, work backwards to get to x, here we have 4 more than 6 times x, so subtract the 4 from both sides...

6x + 4 = 45 becomes

6x = 41

Now get rid of the 6 by dividing both sides by 6 to isolate x...

6x = 41 becomes

x = 41/6 or 6.8333333 as a decimal

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How do you do this question?

Andrei [34K]

Answer:

Solution : Option B, or 9π

Step-by-step explanation:

We are given that y = x, x = 3, and y = 0.

Now assume we have a circle that models the given information. The radius will be x, so to determine the area of that circle we have πx². And knowing that x = 3 and y = 0, we have the following integral:

$\int _0^3$

So our set up for solving this problem, would be such:

$\int _0^3x^2\pi \:$

By solving this integral we receive our solution:

$\int _0^3x^2\pi dx,\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> \pi \cdot \int _0^3x^2dx\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\=> \pi \left[\frac{x^{2+1}}{2+1}\right]^3_0\\=> \pi \left[\frac{x^3}{3}\right]^3_0\\\mathrm{Compute\:the\:boundaries}: \left[\frac{x^3}{3}\right]^3_0=9\\\mathrm{Substitute:9\pi }$

As you can tell our solution is option b, 9π. Hope that helps!

5 0

3 years ago

Read 2 more answers

Simplify the expression below. (–7x + 4) – (2x – 8)

Elanso [62]

Answer:

- 9x + 12

Step-by-step explanation:

Given

(- 7x + 4) - (2x - 8) ← distribute parenthesis by 1 and - 1

= - 7x + 4 - 2x + 8 ← collect like terms

= - 9x + 12

5 0

3 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

densk [106]

Answer:

$\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}$

Step-by-step explanation:

We have:

$\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}$

For sine use:

$\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x$

Substitute:

$\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}$