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2 years ago

Write an equation of the line that passes through (1,1) and is perpendicular to the line that passes through (−6,−5) and (0,7).

Mathematics

1 answer:

iris [78.8K]2 years ago

7 0

Answer:

hope this helps

the long picture will be part 1 and the small pic is the second part

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Find mHK.<br> pls hurry and answer lol

Rashid [163]

It would be mark N? I think

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3 years ago

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What is the midpoint of the line segment with endpoints (3.2, 2.5) and<br> (1.6, -4.5)

jok3333 [9.3K]

Answer:

(2.4, -1)

Step-by-step explanation:

Using midpoint formula plug in the info (x1+x2)/2, (y1+y2)/2

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3 years ago

2u^2 + v when u=3 and v=7

Mariana [72]

Answer: 25

Step-by-step explanation:

There are two variables which is u and v and is given that u is 3 and v is 7 so input that into the expression and solve.

2(3)^2 + 7 Solve the exponent first

2(9) + 7 Now multiply then add

18 + 7 = 25

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3 years ago

7x + 2y = 24<br>8x + 2y = 30

bagirrra123 [75]

(8x + 2y) - (7x + 2y) = 30 - 24

X = 6

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3 years ago

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Find the sumofthe geometrical progression of five terms, of which the first term is 7 and the multiplier is 7.Verify that the su

777dan777 [17]

Answer:

The sum of first five term of GP is 19607.

Step-by-step explanation:

We are given the following in the question:

A geometric progression with 7 as the first term and 7 as the common ration.

$a, ar, ar^2,...\\a = 7\\r = 7$

$7, 7^2, 7^3, 7^4...$

Sum of n terms in a geometric progression:

$S_n = \displaystyle\frac{a(r^n - 1)}{(r-1)}$

For sum of five terms, we put n= 5, a = 7, r = 7

$S_5 = \displaystyle\frac{7(7^5 - 1)}{(7-1)}\\\\S_5 = 19607$

The sum of first five term of GP is 19607.

Verification:

$2801\times 7 = 19607$

Thus, the sum is equal to product of 2801 and 7.

7 0

3 years ago