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forsale [732]
3 years ago
14

I really need help with this my deadline is very soon​

Mathematics
1 answer:
Harman [31]3 years ago
8 0

Answer:

(3)\ y = -x

(5)\ y = -3x -2

Step-by-step explanation:

Required

The equation in slope intercept form

Solving (3):

m = -1

(x_1,y_1) =(4,-4)

The equation in slope intercept form is:

y = m(x - x_1) + y_1

This gives:

y = -1(x - 4) -4

Expand

y = -x + 4 -4

y = -x

Solving (5):

m = -3

(x_1,y_1) = (1,-5)

The equation in slope intercept form is:

y = m(x - x_1) + y_1

This gives:

y = -3(x - 1) - 5

Expand

y = -3x + 3 - 5

y = -3x -2

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(25+3) - 7 x 3 + (2x4)
arlik [135]

Answer:

It is not correct, the answer is 15

Step-by-step explanation:

They forgot the rule of PEMDAS, and accidentally subtracted before they multiplied. It should've gone

(25+3) - 7 x 3 + (2x4)

28-7x3+8

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3 years ago
Need help please :)
swat32

Answer:

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Step-by-step explanation:

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3 years ago
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What is the area of the trapezoid with height 10 units?
Ainat [17]
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5 0
3 years ago
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How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?
irga5000 [103]
The equivalence n \equiv 5 \pmod{12}

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5


for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)


we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009


The last k which produces n<5000 is 416


For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.


Answer: 417

6 0
4 years ago
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