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Rus_ich [418]
3 years ago
7

Can you please answer the question?

Mathematics
1 answer:
Akimi4 [234]3 years ago
5 0
What grade is this and don’t they think you don’t know how to do this
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sukhopar [10]
If he wants to loose 30 points total, subtract what he lost already, 24, from 30. He needs to lose 6 more pounds.
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3 years ago
Slope of the line 4x+6y=12
wolverine [178]
The slope/ gradient of the line is -2/3 or answer c
4 0
3 years ago
Read 2 more answers
Suppose A, R, and B are collinear on AB, and AR:AB= 1/5. What are the coordinates of R?
AfilCa [17]

Answer:

R(1.6, 0)

Step-by-step explanation:

Use the formula, (x, y) = (x_1 + k(x_2 - x_1), y_1 + k(y_2 - y_1)) to find the coordinates of point R, that partition the lie segment AB into the ratio 1/5.

Let,

A(1, -1) = (x_1, y_1)

B(4, 4) = (x_2, y_2)

Thus, plug in the values as follows:

R(x, y) = (1 + \frac{1}{5}(4 - 1), -1 + \frac{1}{5}(4 -(-1))

R(x, y) = (1 + \frac{1}{5}(3), -1 + \frac{1}{5}(4 + 1)

R(x, y) = (1 + \frac{1}{5}(3), -1 + \frac{1}{5}(5)

R(x, y) = (1 + \frac{3}{5}, -1 + \frac{5}{5})

R(x, y) = (\frac{5 + 3}{5}, -1 + 1)

R(x, y) = (\frac{8}{5}, 0)

R(x, y) = (1.6, 0)

The coordinates of point R, are (1.6, 0)

7 0
4 years ago
Can someone please answer this!!!!
alexandr402 [8]

Step-by-step explanation:

\frac{y}{20 \sqrt{3} }  =  \sin(30)

y = 20 \sqrt{3}  \times   \frac{1}{2}

y = 10 \sqrt{3}

\frac{x}{20 \sqrt{3} }  =  \cos(30)

x = 20 \sqrt{3}  \times  \frac{ \sqrt{3} }{2}

x = 30

option D, x=30 and y = 10√3

3 0
3 years ago
the percent of fat calories that a person in america consumes each day is normally disributed with a mean of about 36 and a stan
Ugo [173]

Answer:

True

Step-by-step explanation:

Given that

The mean is 36

And, the standard deviation is 10

We take the sample of 16 individuals and the same would be randomly chosen

We need to find out that whether the given statement is true or false

So,

The given statement is true as the random variable means the average percent that could be consumed each day by a person

8 0
3 years ago
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