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Archy [21]
4 years ago
7

Find the equation of a line, in slope-intercept form of a line that passes through the point (-5, -1) and is parallel to the lin

e -2x + 4y = 8.
Mathematics
2 answers:
MrMuchimi4 years ago
5 0

Answer:

y=(1/2)x +(3/2)

Step-by-step explanation:

aksik [14]4 years ago
3 0
First you need he equation in y=mx+b so that you can see the slope. then you take the (-5,-1) and plug them in for x and y. then solve for b. Then put the answer to b into the y=mx+b...parallel means they have the same slope.

y=(1/2)x +2
now using the (-5,-1)
-1=(1/2)(-5) +b
b=(3/2)

The new equation is y=(1/2)x +(3/2). This is parallel and passes through (-5,-1).
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Find the perimeter of the triangle whose vertices are the following, specified points in the plane.
hichkok12 [17]

Given:

The vertices of the triangle are (-10,-3), (1,4) and (-1,7).

To find:

The perimeter of the triangle.

Solution:

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let the vertices of the triangle are A(-10,-3), B(1,4) and C(-1,7).

Using distance formula, we get

AB=\sqrt{(1-(-10))^2+(4-(-3))^2}

AB=\sqrt{(1+10)^2+(4+3)^2}

AB=\sqrt{(11)^2+(7)^2}

AB=\sqrt{121+49}

AB=\sqrt{170}

Similarly,

BC=\sqrt{\left(-1-1\right)^2+\left(7-4\right)^2}=\sqrt{13}

AC=\sqrt{\left(-1-\left(-10\right)\right)^2+\left(7-\left(-3\right)\right)^2} =\sqrt{181}

Now, the perimeter of the triangle is

Perimeter=AB+BC+AC

Perimeter=\sqrt{170}+\sqrt{13}+\sqrt{181}

Perimeter\approx 13.038+3.606+13.454

Perimeter=30.098

Therefore, the perimeter of the triangle is 30.098 units.

8 0
3 years ago
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Step-by-step explanation:

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