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Dennis_Churaev [7]
1 year ago
10

which equation represents a line that is parallel to the line whose equation is 3x-2y=7; which equation represents a line parall

el to the line shown on the graph?; which equation represents a line that is parallel to the line whose equation is y=-3x; which equation represents a line that is perpendicular to the line represented by 2x-y=7; which equation represents a line that is parallel to the line y=-3x+4; which equation represents a line that is parallel to the line y=3-2x; which equation represents a line that is parallel to the line y=-4x+5; which equation represents a line that is perpendicular to the graph of y 2 5x 1
Mathematics
1 answer:
aleksley [76]1 year ago
3 0

The equation of the line that is parallel to the line whose equation is 3x-2y=7 would be y = 3/2x + b, in which b can be any real number.

How are parallel straight lines related?

Parallel lines have the same slope since the slope is like a measure of steepness and since parallel lines are of the same steepness, thus, are of the same slope.

We have been given a parallel line with has equation

3x-2y=7

In order to solve this, the slope of the original line.

3x - 2y = 7

-2y = -3x + 7

y = 3/2x - 7/2

thus its slope is 3/2.

thus, the slope of the needed line is 3/2 too.

we know that any line that is parallel to that would have this slope.

So anything is written in the form:

y = 3/2x + b

The equation of the line that is parallel to the line whose equation is 3x-2y=7 would be y = 3/2x + b, in which b can be any real number.

Learn more about parallel lines here:

brainly.com/question/13857011

#SPJ4

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A

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Because 5x6=30

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In the right ∆ABC, the hypotenuse AB = 17 cm. M is the midpoint of the hypotenuse. Find the legs if PAMC=32 cm and PBMC=25 cm
jeyben [28]

Answer:

The length of the legs is 8.64cm and 14.64cm respectively

Step-by-step explanation:

I've added an attachment to aid my explanation.

At different intervals, I'll be making reference to it.

Given

AB = 17

PAMC = 32

PBMC = 25

From the attachment, we have:

y + z = AB

Since, M is the Midpoint

y = z = \½AB

Substitute 17 for AB

y = z = \½ * 17

y = z = 8.5

Also, from the attachment

v + x + z = PAMC

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Substitute 8.5 for y

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v + x = 23.5 --------- (1)

Also, from the attachment

v + w + z = 25

Substitute 8.5 for z

v + w + 8.5 = 25

v + w = 25 - 8.5

v + w = 17.5 ----------- (2)

Subtract (2) from (1)

v - v + x - w = 23.5 - 17.5

x - w = 6

Make x the subject

x = 6 + w

Apply Pythagoras Theorem:

We have that:

AB^2 = AC^2 + BC^2

The above can be replaced with

17^2 = x^2 + w^2 (see attachment)

289 = x^2 + w^2

Substitute 6 + w for x

289 = (6 + w)^2 + w^2

289 = 36 + 12w + w^2 + w^2

289 - 36 = 12w + 2w^2

253 = 12w + 2w^2

Reorder

2w^2 + 12w - 253 = 0

Solve using quadratic equation:

w = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

Where

a = 2

b = 12

c = -253

w = \frac{-12 \± \sqrt{12^2 - 4 * 2 * -253}}{2 * 2}

w = \frac{-12 \± \sqrt{144 + 2024}}{4}

w = \frac{-12 \± \sqrt{2168}}{4}

w = \frac{-12 \± 46.56}{4}

Split:

w = \frac{-12 + 46.56}{4} or w = \frac{-12 - 46.56}{4}

w = \frac{34.56}{4} or w = \frac{-58.56}{4}

w = 8.64 or w = -14.64

But length can't be negative

So:

w = 8.64

Recall that: x = 6 + w

x = 6 + 8.64

x = 14.64

<em>Hence, the length of the legs is 8.64cm and 14.64cm respectively</em>

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