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fredd [130]
3 years ago
8

What is 1 3/10 • 1 1/3

Mathematics
1 answer:
Komok [63]3 years ago
8 0

Answer:

Step-by-step explanation:

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Sofia lives in a $105,000 brick house in the suburbs with contents valued at
Zinaida [17]

Answer:$58.88

Step-by-step explanation:

6 0
3 years ago
Describe the type of solution for the given system of linear equations.
Kazeer [188]

Answer:

B

Step-by-step explanation:

5 0
3 years ago
A video game store allows customers to rent games for $4.75 each. Customers can also buy a membership for $54 each year, and vid
Nezavi [6.7K]

Answer:

24

Step-by-step explanation:

Cost of membership + game = 2.50x + 54

Cost of just games = 4.75x

The catch is, how do we make up the 54 dollars?

So equate the 2 costs

4.75x = 2.50x + 54      Subtract 2.50

4.75x - 2.50x = 54       Combine

2.25x = 54                    Divide by 2.25

x = 54/2.25

x = 24

That means that 24 is the break even point.

5 0
3 years ago
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

3 0
2 years ago
Quadrilateral ABCD has vertices A(1, -3) , B(2, 1), C(6, 3), and D(7, -1). Prove or disprove that the Quad ABCD is a rectangle.
zavuch27 [327]
If AC=BD, than Quad ABCD is a rectangle
AC
√(3+3)²+(6-1)²
√6²+5²
√36+25
√61
BD
√(1+1)²+(2-7)²
√2²+(-5)²
√4+25
√29
Quad ABCD is not a rectangle because √61 and √29 are not equal.
7 0
3 years ago
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