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Nadya [2.5K]
3 years ago
7

Which option correctly shows how this formula can be rearranged to isolate x^2?

Mathematics
1 answer:
Dimas [21]3 years ago
4 0

Answer:

A

Step-by-step explanation:

We have:

\displaystyle m=\frac{x_1-x_2}{y_1-y_2}

And we want to isolate x₂.

So, let’s first remove the denominator by multiplying both sides by it:

\displaystyle m(y_1-y_2)=\frac{x_1-x_2}{y_1-y_2}(y_1-y_2)

The right side will cancel. This will leave:

\displaystyle m(y_1-y_2)=x_1-x_2

Now, we can subtract x₁ from both sides. So:

\displaystyle m(y_1-y_2)-x_1=-x_2

Finally, we will multiply both sides by -1. So:

\displaystyle x_2=-m(y_1-y_2)+x_1

Hence, our answer is A.

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Expand: x(3 + 9). please
Oksi-84 [34.3K]

Answer:

x(3 + 9) = 3x + 9x = 12x

Step-by-step explanation:

...........

8 0
3 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
Find the area of each figure. Round to the nearest tenth if necessary.
Tanzania [10]

Answer:

64 cm²

Step-by-step explanation:

(12*4.5)+(2*5)= 54+10=64

6 0
3 years ago
A model giraffe has a scale of 1 in : 3 ft. If the model giraffe is 4 in tall, then how tall is the real giraffe
VladimirAG [237]
The answer is 12ft. Because if 1in is 3ft then 4in will be 4 multiplied by 3
4 0
3 years ago
Read 2 more answers
Let’s assume the 147 freshman maintained a 65% attendance rate during the first 4 week of school. If our next 4 week goal for at
shutvik [7]

Answer:

  • 12 students  

Step-by-step explanation:

If 65% equals 147

Then to maintain 70% there is a need for 5% more

<u>Which equals</u>

  • 5*147/65 ≈ 12 students
4 0
3 years ago
Read 2 more answers
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