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ZanzabumX [31]
3 years ago
12

3/9=21/k what is the strategy for solving it

Mathematics
1 answer:
Vinvika [58]3 years ago
5 0

Answer: k = 63

Step-by-step explanation: You need to multiply the left side by its reciprocal to move the numbers to the right. You also need to multiply by k on the right side to get it alone on the left. This will result in

(9/3)(3/9) = (21/k)(9/3) simplified

1 = 189/3k • k    then multiply by k

k = 189/3 simplify

k = 63

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2+d=2-3(d-5)-2 Pls show work
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2 + d = 2 - 3(d- 5) - 2

First, cancel 2 on both sides. / Your problem should look like: d = -3(d - 5) - 2
Second, expand. / Your problem should look like: d = -3d + 15 - 2
Third, simplify -3d + 15 - 2 to get -3d + 13. / Your problem should look like: d = -3d + 13
Fourth, add 3d to both sides. / Your problem should look like: d + 3d = 13
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3 0
4 years ago
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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
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Answer:

66.42°

Step-by-step explanation:

<u>Given:</u>

A(5, 3t+2, 2)

B(1, 3t, 2)

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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)

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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}

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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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