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3 years ago

Given that y = 18 + 3x² i) express x in terms of y

Mathematics

1 answer:

fgiga [73]3 years ago

6 0

Answer:

$\sqrt{-6 + \frac{y}{3}} = x$

Step-by-step explanation:

y = 18 + 3x²

- 18 - 18

__________

$\frac{-18 + y}{3} = \frac{3x^2}{3} \\ \\ -6 + \frac{y}{3} = x^2 \\ \\ \sqrt{-6 + \frac{y}{3}} = x$

I am joyous to assist you anytime.

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2 years ago

A stem-and-leaf plot is made using the following values. Which does NOT correctly represent a line of stem and leaves from the p

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2 years ago

Read 2 more answers

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

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2 years ago

What number is between 3.16 and 3.17?

VladimirAG [237]

There are multiple answers to this question
Answers: 3.161, 3.162, 3.163, 3.164, 3.165, 3.166, 3.167, 3.168, 3.169
(You can just pick one)

Reason: just look at the place numbers. The number has to be between 3.16 and 3.17. So go to the next place value which is the thousandths place. 3.161-3.169 all are between since they have a 6 in the hundredths place, but does not go over 1.17

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3 years ago

What is the answer to 2(x+1)<x+3

Temka [501]

2x+2 < x+3
2x-x < 3-2
x <1

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2 years ago