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2 years ago

Find two positive numbers whose difference is 9 and whose product is 2950.

Mathematics

1 answer:

Naddika [18.5K]2 years ago

5 0

The two positive numbers whose difference is 9 and whose product is 2950 are 50 and 59

<h3>How to determine the positive numbers?</h3>

As a general rule, it should be noted that positive numbers are numbers that have their value greater than 0

So, we start by representing the two positive numbers with x and y.

So, we have the following equations

x - y = 9

xy = 2950

Make x the subject in the first equation x - y = 9

x = y + 9

Substitute y + 9 for x in the second equation

(y + 9) * y = 2950

Expand the equation

y^2 + 9y - 2950 = 0

Using a graphing tool, we have the solution of the above equation to be

y = 50

Recall that:

x = 9 + y

So, we have:

x = 9 + 50

Evaluate

x = 59

Hence, the two positive numbers whose difference is 9 and whose product is 2950 are 50 and 59

Read more about positive numbers at:

brainly.com/question/1782403

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Step-by-step explanation:

solution:

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3 years ago

Find the area of each parallelogram. What is the relationship between the areas?

castortr0y [4]

Answer:

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In parallelogram TQRS.

Coordinate of TQRS are;

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Coordinate of T'Q'R'S' are;

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Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

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Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

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