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3 years ago

Is(2, 5)asolutiontotheinequality 2x + 3y ≥ 20 ?

Mathematics

2 answers:

Daniel [21]3 years ago

5 0

Answer:

Step-by-step explanation:

2x + 3y ≥ 20

Substitute the point (2,5) in for x and y and see the inequality is true

2(2) +3(5) ≥ 20

4 + 15 ≥ 20

19 ≥ 20

This is not true, so (2,5) is not a solution

Inessa05 [86]3 years ago

3 0

Steps to solve:

2x + 3y ≥ 20; x = 2, y = 5

~Substitute

2(2) + 3(5) ≥ 20

~Simplify

4 + 15 ≥ 20

~Simplify

19 ≥ 20

The inequality is NOT true. Therefore, (2,5) is NOT a solution to the inequality.

Best of Luck!

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Answer:

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Step-by-step explanation:

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2 years ago

A street light is at the top of a 25 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 6

Burka [1]

Answer:

Tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

Step-by-step explanation:

Given : In the figure attached, Length of girl EC = 4 ft

Length of street light AB = 25 ft

Girl is moving away from the light with a speed = 6 ft per sec.

To Find : Rate ( $\frac{dw}{dt}$ ) of the tip (D) of the girl's shadow (BD) moving away from th

light.

Solution : Let the distance of the girl from the street light is = x feet

Length of the shadow CD is = y feet

Therefore, $\frac{dx}{dt}=6$ feet per sec. [Given]

In the figure attached, ΔAFE and ΔADE are similar.

By the property of similar triangles,

$\frac{x}{21}=\frac{x+y}{25}$

25x = 21(x + y)

25x = 21x + 21y

25x - 21x = 21y

4x = 21y

y = $\frac{4x}{21}$

Now we take the derivative on both the sides,

$\frac{dy}{dt}=\frac{4}{21}\times \frac{dx}{dt}$

= $\frac{4}{21}\times 6$

= $\frac{8}{7}$

≈ 1.14 ft per sec.

Since w = x + y

Therefore, $\frac{dw}{dt}= \frac{dx}{dt}+\frac{dy}{dt}$

$\frac{dw}{dt}=6+1.14$

= 7.14 ft per sec.

Therefore, tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

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3 years ago