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2 years ago

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The altitude (i.e., height) of a triangle is increasing at a rate of 1.5 cm/minute while the area of the triangle is increasing

at a rate of 3 square cm/minute. at what rate is the base of the triangle changing when the altitude is 7.5 centimeters and the area is 99 square centimeters?

1 answer:

Stolb23 [73]2 years ago

6 0

99+7.5 +3 +1.5 =111 that your awner 111

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2 years ago

When Carson runs the 400 meter dash, his finishing times are normally distributed with a mean of 63 seconds and a standard devia

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Answer: in 95% of races, his finishing time will be between 62 and 64 seconds.

Step-by-step explanation:

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below

68% of data falls within the first standard deviation from the mean.

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2 standard deviations = 2 × 0.5 = 1

63 - 1 = 62 seconds

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2 years ago

What is the surface area of the cylinder with height 2 yd and radius 3 yd? Round your answer to the nearest thousandth.

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<h2>Given :</h2>

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2 years ago

Determine whether a figure with the given vertices is a rectangle using the Distance Formula.

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Answer:

Yes; Opposite sides are congruent, and diagonals are congruent.

Step-by-step explanation:

we have

$A(4, -7), B(4, -2), C(0, -2), D(0, -7)$

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the length of the sides

<u><em>Find the distance AB</em></u>

substitute the values

$d=\sqrt{(-2+7)^{2}+(4-4)^{2}}$

$d=\sqrt{(5)^{2}+(0)^{2}}$

$AB=5\ units$

<u><em>Find the distance BC</em></u>

substitute the values

$d=\sqrt{(-2+2)^{2}+(0-4)^{2}}$

$d=\sqrt{(0)^{2}+(-4)^{2}}$

$BC=4\ units$

<u><em>Find the distance CD</em></u>

substitute the values

$d=\sqrt{(-7+2)^{2}+(0-0)^{2}}$

$d=\sqrt{(-5)^{2}+(0)^{2}}$

$CD=5\ units$

<u><em>Find the distance AD</em></u>

substitute the values

$d=\sqrt{(-7+7)^{2}+(0-4)^{2}}$

$d=\sqrt{(0)^{2}+(-4)^{2}}$

$AD=4\ units$

Compare the length sides

AB=CD

BC=AD

therefore

Opposite sides are congruent

step 2

Find the length of the diagonals

<u><em>Find the distance AC</em></u>

substitute the values

$d=\sqrt{(-2+7)^{2}+(0-4)^{2}}$

$d=\sqrt{(5)^{2}+(-4)^{2}}$

$AC=\sqrt{41}\ units$

<u><em>Find the distance BD</em></u>

substitute the values

$d=\sqrt{(-7+2)^{2}+(0-4)^{2}}$

$d=\sqrt{(-5)^{2}+(-4)^{2}}$

$BD=\sqrt{41}\ units$

Compare the length of the diagonals

AC=BD

therefore

Diagonals are congruent

The figure is a rectangle, because Opposite sides are congruent, and diagonals are congruent

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