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3 years ago

If L is the line through the point (-3,5) having a slope of (2/3) complete the equation of L

1 answer:

4 0

Y = mx + b
slope(m) = 2/3
(-3,5)...x = -3 and y = 5
now sub and find b, the y int
5 = 2/3(-3) + b
5 = -2 + b
5 + 2 = b
7 = b

so ur equation is : y = 2/3x + 7

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4 years ago

Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if

Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

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

Using distance formula, we get

$JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}$

$JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}$

$JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}$

$JK=\sqrt{4+9}$

$JK=\sqrt{13}$

Similarly,

$KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6$

$LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}$

$JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6$

Now, perimeter P of ▱JKLM is

$P=JK+KL+LM+JM$

$P=\sqrt{13}+6+\sqrt{13}+6$

$P=2\sqrt{13}+12$

$P=2(3.61)+12$

$P=7.22+12$

$P=19.22$

$P\approx 19.2$

Therefore, the perimeter P of ▱JKLM is 19.2 units.

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3 years ago

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fiasKO [112]

Answer:

1/3 ( 5y-2) = x

Step-by-step explanation:

5y = 3x+2

Subtract 2 from each side

5y -2 = 3x+2-2

5y-2 = 3x

Divide each side by 3

1/3 ( 5y-2) = 3x/3

1/3 ( 5y-2) = x

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3 years ago

For each value of u, determine whether it is a solution to 7 > u.

Nezavi [6.7K]

7 > u is the same as u < 7. So, any number, integer, fraction or mixed number, that is smaller than 7, is a solution to 7>u.

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3 years ago

How do you write 103,727,495 in expanded form?

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