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Tpy6a [65]
3 years ago
7

List all the pairs of integers with a product of 36. then find the pair whose sum is 20

Mathematics
2 answers:
strojnjashka [21]3 years ago
5 0

Answer: It must be (18,2) or (2,18).

Step-by-step explanation:

Since we have given that

The product of pairs of integers = 36

So, pairs must be

(1,36)

(2,18)

(3,12)

(4,9)

(6,6)

(9,4)

(12,3)

(18,2)

(36,1)

But we need the pair of integers whose sum is 20.

So, it must be (18,2) or (2,18).

RoseWind [281]3 years ago
4 0
This is an excellent practice for the solution of quadratic equations.

1*36=36  => (1,36)
2*18=36  => (2,18)
3*12=36  => (3,12)
4*9=36   => (4,9)
6*6=36   => (6,6)
9*4=36   => (9,4)
12*3=36  => (12,3)
18*2=36  => (18,2)
36*1=36  => (36,1)

We can see that the sum decreases until the two factors are close (or equal) and then increases again.

The pair of integers with a sum of 20 is therefore (2,18) or (18,2).
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triangle ABC above is isosceles with AB=AC and BC =48. the ratio of DE is 5:7. what is the length of DC
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Answer:

28

Step-by-step explanation:

Question Image is attached.

Triangle BED and Triangle CFD are similar.

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5x + 7x = 48

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x = 4

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Complete the point-slope equation of the line through (1,3) (5,1) y-3=?
MaRussiya [10]

Answer:

\huge\boxed{y-3=-\frac{1}{2}(x-1)}

Step-by-step explanation:

Point-slope is:

y-y_1=m(x-x_1)

m-\text{This represents the slope.}\\\\(x_1,y_1)-\text{This represents the point used in the equation.}

<h2>Our goal: </h2>

We have to complete the point-slope equation of the line through (1,3) (5,1).

--------------------------------------------------------

We have a incomplete equation of the line.

y-3=m(x-x_1)

We need to find the <u>slope</u> of the line, and the <u>value</u> of  x_1.

--------------------------------------------------------

<h3>Finding 'x1':</h3>

It seems that the value of 3 was used to be y_1. This means that the point (1,3) was used for the equation. This means that x_1 would have to be 1.

<h3>Finding Slope:</h3>

Slope is rise over run.

m=\frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}

We are given the points (1,3) and (5,1).

m=\frac{1-3}{5-1}=\frac{-2}{4}=\frac{-1}{2}=\boxed{-\frac{1}{2}}

The slope is one-half.

--------------------------------------------------------

We now have enough information to complete the point-slope equation.

{\left \{ {{x_1=1} \atop {m=-\frac{1}{2} }} \right.}\\\\y-3=m(x-x_1)\rightarrow\boxed{y-3=-\frac{1}{2}(x-1)}

Our final equation is:

y-3=-\frac{1}{2}(x-1)

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