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krek1111 [17]
3 years ago
5

Write an equation of the line perpendicular to 6x + 7y=126, through (7,8).​

Mathematics
1 answer:
Mandarinka [93]3 years ago
6 0

Answer:

y = 7/6 * x - 1/6

Step-by-step explanation:

6x + 7y = 126

y = (- 6/7) * x  + 18

slope: -6/7

A line perpendicular to another has a slope that is the negative reciprocal of the slope of the other line

slope' : 7/6

y = mx + b

8 = 7/6 * 7 + b

b =  (48-49) / 6 = - 1/6

equation: y = 7/6 * x - 1/6

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2 years ago
Juan pays a one-time fee of $30.
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Answer:

Step-by-step explanation:

A. c=30+45(h)

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1 year ago
Find the length of QP given Q is the midpoint of XF, PQ =2X+1, XF=7X-4, PF=X
nadezda [96]

Answer:

Therefore the length of QP = 3.4 units

Step-by-step explanation:

Given:

PQ = 2x + 1

XF = 7x - 4

PF = x

Q is the mid poimt of XF

∴ XQ = QF

QF = PQ - PF  ..........( Q - F - P )

     = 2x + 1 - x

∴ QF      = x + 1

∴ XQ = QF = x + 1

TO Find:

QP = ?

Solution:

By Addition Property we have

XP = XQ + QF+PF ..........(X-Q-F-P)\\\\

XF + PF =XQ + QF+PF ..........(X-Q-F-P)\\

Substituting the given values in above equation we get

(7x - 4) + x = (x +1) + (x +1) + x

8x -4 = 3x +2

8x - 3x + 4 + 2

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∴ x = \frac{6}{5}

Now we require

QP = (2x + 1)

∴ QP = 2\times \frac{6}{5} +1\\\\QP = \frac{12+5}{5} \\\\QP =\frac{17}{5} \\\\\therefore QP = 3.4\ unit

Therefore the length of QP = 3.4 units

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3 years ago
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