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ludmilkaskok [199]
3 years ago
7

12-2/3x=x-18 so how do you answer this question

Mathematics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

Step-by-step explanation:

I think I've already answered this question.

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Find an equation of the line that passes through the pair of points (-5, -9) and (2,-7)
vladimir1956 [14]

Answer:

y=0.3x - 7.6

Step-by-step explanation:

line: y=ax+b

a= Δy ÷ Δx

a= (-9 + 7) ÷ (-5 - 2)

a= -2 ÷ -7 = approximately 0.3

y=0.3x + b

Now you can fill in one of the points to find b (or both to make sure you got the answer right).

0.3 × 2 + b = -7

0.6 + b = -7

b = -7 - 0.6 = approximately -7.6

So: an equation of the line that passes through this pair of points is y=0.3x - 7.6

On a test, i would write as many decimals as possible if not said not to, to get the most exact answer. Also remember to keep calculating with the number on your calculator and not to round them off for the most exact answer.

4 0
3 years ago
Work out the area of a parallelogram of base 5m and height 3m
Arada [10]
B*H= area so 3*5= 15

15 is your answer
6 0
3 years ago
Find the areas of the trapezoids.
Juliette [100K]

Answer:

12 square units

Step-by-step explanation:

A = 1/2·h(sum of bases)

A = 1/2(4)(2+4)

A = 2(6) which is 12 square units

3 0
3 years ago
write the equation for the nth term of the sequence and then find the value of the 24th term: 4, 13, 22, 31, ...
julia-pushkina [17]

Answer :

nth term

an = a1 + (n-1) d

an = 4 + (9-1) 9

an = 4 + (8)9

an = 4 + 72

an = 82

24th term

an = 4 + (24-1) 9

an = 4 + (23) 9

an = 4 + 207

an = 211 is the answer

Hope it helped

4 0
2 years ago
Using an ordered alphabet of 26 letters, how many ways are there to choose a set of six letters such that no two letters in the
mestny [16]
To find our solution, we can start off by creating a string of 27 boxes, all followed by the letters of the alphabet. Underneath the boxes, we can place 6 pairs of boxes and 15 empty boxes.The stars represent the six letters we pick. The empty boxes to the left of the stars provide the "padding" needed to ensure that no two adjacent letters are chosen. We can create this - 
\binom {21} {6}= \frac{21*20*19*18*17*16}{6*5*4*3*2*1} =21*19*17*8=54264
Thus, the answer is that there are \boxed{54264} ways to choose a set of six letters such that no two letters in the set are adjacent in the alphabet. Hope this helped and have a phenomenal New Year! <em>2018</em>
5 0
3 years ago
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