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AnnZ [28]
4 years ago
5

g The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2

/min. At what rate is the base of the triangle changing when the altitude is 20 cm and the area is 160 cm2
Mathematics
1 answer:
horsena [70]4 years ago
5 0

Answer:

  -0.6 cm/min

Step-by-step explanation:

The formula for the area of a triangle is ...

  A = (1/2)bh

Solving for the base, we find ...

  b = 2A/h

Then the rate of change of the base is ...

  b' = 2(A'h -Ah')/h^2

Filling in the given values, we find the rate of change of the base to be ...

  b' = 2((2 cm^2/min)(20 cm) -(160 cm^2)(1 cm/min))/(20 cm)^2

  = 2(40-160)/400 cm/min

  = -0.6 cm/min

The base is decreasing at 0.6 cm/minute.

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Answer:

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Step-by-step explanation:

It is because in a span of 1 x coordinate to the right it went up by 6 so it is 6 over one in fraction form which is 6.

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Sharon uses 72 centimeters of ribbon to wrap gifts.she uses 24 centimeters of her total ribbon to wrap a big gift she use the re
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She will use 8 centimeters of ribbon on each small gift.

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Consider the points A(5, 3t+2, 2), B(1, 3t, 2), and C(1, 4t, 3). Find the angle ∠ABC given that the dot product of the vectors B
Vilka [71]

Answer:

66.42°

Step-by-step explanation:

<u>Given:</u>

A(5, 3t+2, 2)

B(1, 3t, 2)

C(1, 4t, 3)

BA • BC = 4

Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

BC = OC - OB = (1-1, 4t-3t, 3-2) = (0, t, 1)

Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

2 = t

t = 2

Now we know that:

BA = (4, 2, 0)

BC = (0, 2, 1)

Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

BA • BC = 4

|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}

|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}

2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4

10\cos{(m\angle{ABC})} = 4

\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}

m\angle{ABC} = cos^{-1}{(\frac{2}{5})}

m\angle{ABC} = 66.4218^{\circ}

Rounded to two decimal places:

m\angle{ABC} = 66.42^\circ

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2 years ago
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m - 2/3 + 1/4 = 2/12

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What is 2/3 of $300?
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Well one third is 100
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