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Anika [276]
3 years ago
14

The sum of 2 numbers (x and y) is 20, and their product is 51. What are the values of x and y

Mathematics
1 answer:
Murljashka [212]3 years ago
4 0

Answer: 13&7

Step-by-step explanation:

The sum of two numbers x&y is 20

Product of the 2numbers xy is 51

X+y=20......equation 1

XY=51.........equation 2

X=20-y.......equation 3

Substitute for x inequation 3 into equation 2

So we have;

(20-y)y=51

Open the bracket

20y-y^2=51

Collect like terms,so we have

Y2-20y+51=0

Factorise

Factors of 51 is 13&7

So -13-7=-20 &-13×-7=51

Substitute

Y2-13y-7y+51=0

Y(y-13)-7(y-13)=0

(Y-7)(y-13)=0

Y-7=0

Y=7 or

Y-13=0

Y=13

Y=13 or 7

Substitute for y in equation 1

X+y=20

X+7=20

X=20-7

X=13 or

X+13=20

X=20-13

X=7

Therefore, X &Y = 13 &7

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Using translation concepts, it is found that the transformations needed to form g(x) = 3(x + 2)² - 1 from f(x) = x² are given as follows:

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<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the change is as follows:

f(x) = x² -> g(x) = 3(x + 2)² - 1.

f(x) is multiplied by 3 to form g(x), hence g(x) has a narrower graph. Since x -> x + 2, it was shifted left 2 units, and since y -> y - 1, it was shifted down 1 unit.

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2 years ago
Find the distance from point $A\left(15,-21\right)$ to the line $5x+2y\ =\ 4$ . Round your answer to the nearest tenth.
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Given the coordinate (15, -21) and the line 5x + 2y = 4

In order to get the point on the line 5x + 2y =4, we can a point on the line

Let x = 0

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The point (0, 2) is on the line.

Find the distance between the point (15, -21) and (0, 2) using the distance formula

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Hence the distance from point (15,-21) to the line 5x + 2y = 4 is 27.5 units

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We can designate a variable, x, to represent the width of the rectangle.  To represent the length and widths of the rectangle, we can say that the length is 3x-4 and the width is just x.  We can then write an equation for the perimeter and solve:

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