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Stells [14]
3 years ago
12

Please can some help me in this?

Mathematics
1 answer:
tino4ka555 [31]3 years ago
5 0

Answer:

n = 70

Step-by-step explanation:

14 * x < 80

x = 6 won't work

14 * 6 = 84 which beyond the boundry.

14 * 5 = 70 which is less than 80 and is divisible by 14

14 * 3 = 42

70 does not have a 3 in it.

So the largest number that has 14 in it and is less than 80 is 70

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Question 3<br> Find the equation of the line joining (-2,4) and (3,7)
otez555 [7]

Answer:

y = (3/5)x+26/5 or 5y = 3x+26

Step-by-step explanation:

Applying,

The equation of a line in two point form

(y₂-y₁)/(x₂-x₁) = (y-y₁)/(x-x₁)............... Equation 1

From the question,

Given: y₂ = 7, y₁ = 4, x₂ = 3, x₁ = -2

Substitute these values into equation 1

(7-4)/[3-(-2)] = (y-4)/(x+2)

3/5 = (y-4)/(x+2)

5(y-4) = 3(x+2)

5y-20 = 3x+6

5y = 3x+6+20

5y = 3x+26

y = (3/5)x+26/5

Hence the equation of the line is y = (3/5)x+26/5 or 5y = 3x+26

7 0
3 years ago
What's 1÷0<br>If you got this right this means you're gifted!!
Olegator [25]
1 ÷ 0 is undefined. nothing can be divided by 0.
3 0
3 years ago
Read 2 more answers
Which line is a linear model for the data?
DochEvi [55]

Answer:

Bottom Left Choice

Step-by-step explanation:

If you look at the points, they are traveling up and to the right. So is the line in the graph. Also, the line in the graph is closer to more dots than the others.

5 0
3 years ago
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What is median of 3 7 5 2 4 2 53 6 9 7 7 3 5 2 7 3 2 2 7
ludmilkaskok [199]

Answer:

5

Step-by-step explanation:

To find the median number of

3 7 5 2 4 2 53 6 9 7 7 3 5 2 7 3 2 2 7

Number them from least to greatest

2 2 2 2 3 3 3 4 5 5 6 7 7 7 7 7 9 53

Since this isn't an even numbered set, all we do is cancel each number out equally on both sides until only one number which should be in the middle is left. This gives you an answer of 5

Hope this helps!

3 0
3 years ago
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The price of an item is reduced 70% the original price $90 what is the price now?
uranmaximum [27]
90 - (90 × .7) = $27

hope this helps
7 0
3 years ago
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