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kaheart [24]
3 years ago
5

Write an equation of the line that passes through (3, 7) and is parallel to the line shown. (1, 4) (- 2, - 2)

Mathematics
1 answer:
-Dominant- [34]3 years ago
3 0

Answer:

y=2x+1.

Step-by-step explanation:

You first need to find the slope of the line already graphed. -2 to 1 would be three units to the right, and -2 to 4 would be six units up. You would use the equation (y2-y1)/(x2-x1), where x1 and y1 would be the x and y values of the first coordinate of the graphed line, and x2 and y2 would be the second coordinate. This would be 6/3, which can be simplified to 2/1, which can be simplified to 2, which would mean the slope of the given line would be 2, as well as the parallel line. Now, you could use a pretty complicated equation to find the y-intercept of the line you need to graph, but I would recommend using the slope to find it. Since the given point is (3, 7), you would need to go to the left, meaning you would reverse the slope until you reach the y-axis, so you would use -2 as the slope. 3*-2=-6, and 7-6=1, so the y-intercept would be 1. The answer would be y=2x+1. Hope this helps.

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Y = x + 2<br> y = 2x - 5<br> what is x and y
lana66690 [7]

Answer:

x= 7 y= 9 (7,9)

Step-by-step explanation:

solve by using system if equations. multiply top answer by 2, giving u 2y=2x+4 and y=2x-5 subtract the equations. y=9 (the x cancels out) then put 9 in for y. 9=x+2. subtract 2. giving u 7. and now x =7

7 0
2 years ago
Which of the following could be the equation of the graph shown below
irina [24]
-4x + 2y = -6
2y = 4x - 6
y = 2x - 3....y int = -3 , x int = (3/2,0)

I believe it is going to be : -4x + 2y = -6

4 0
3 years ago
Read 2 more answers
He has 9 cans of white paint and 20 cans of blue paint.
Liula [17]

Answer:

The answer to the problem is 1/2

3 0
3 years ago
5x-y&gt;4 how to solve this one​
Allushta [10]

If you want me to solve for "Y" then, the answer will be,

y  < 5x - 4

ir if you want me to solve for "X" instead then the answer will be,

x >  \frac{4 + y}{5}

3 0
2 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
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