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3 years ago

P(1, 10)and Q(7,4) are the endpoints of a line segment. What is the midpoint M of that line segment

Mathematics

2 answers:

melamori03 [73]3 years ago

6 0

Answer:

(4, 7)

Step-by-step explanation:

The midpoint formula is (((x1 + x2)/2), ((y1+y2)/2)))

x1 +x2 = 1 +7 = 8

8/2 = 4

y1 + y2 = 10 + 4 = 14

14/2 = 7

(4, 7) is the midpoint

Crazy boy [7]3 years ago

3 0

Answer:

M = (4, 7)

Step-by-step explanation:

Using the midpoint formula

M = [ $\frac{1}{2}$ (x₁ + x₂ ), $\frac{1}{2}$ (y₁ + y₂ ) ]

with (x₁, y₁ ) = P(1, 10) and (x₂, y₂ ) = Q(7, 4)

M = [ $\frac{1}{2}$ (1 + 7), $\frac{1}{2}$ (10 + 4) ] = (4, 7)

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15 points I really need help with this and it’s due soon please help me and explain.

PilotLPTM [1.2K]

A square's diagonal has a length equal to √(2) times the length of its sides. So if the side length is x, then the diagonal is such that

√(2) x = 10 cm

→ x = 10/√(2) cm ≈ 7.07 cm

The perimeter of a square is 4 times its side length, so the perimeter is

4 (10/√(2) cm) = 40/√(2) cm ≈ 28.3 cm

which makes 28.4 cm the closest answer.

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2 years ago

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175 pages in 5 days = __ pages in 2 days

barxatty [35]

In 5 days, = 175 pages
so, in 1 day = 175/5 Pages

Now, In 2 days, it would be: 175/5 * 2 = 35 * 2 = 70

So, your final answer is 70

Hope this helps!

8 0

3 years ago

Read 2 more answers

(7z^3+42z^2-15z+1)- (36z^2+12z-21)

My name is Ann [436]

Answer:

7z^3+6z^2-27z+22

Step-by-step explanation:

The given question is of subtraction:

So,

(7z^3+42z^2-15z+1)- (36z^2+12z-21)

=7z^3+42z^2-15z+1-36z^2-12z+21

Combining alike terms to simplify

=7z^3+42z^2-36z^2-15z-12z+1+21

=7z^3+6z^2-27z+22

So the answer in simplified form is:

7z^3+6z^2-27z+22 ..

6 0

2 years ago

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Please help with math

jeka94

What would you need help with?

4 0

2 years ago

Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, nor

kenny6666 [7]

Answer:

a) 0.0167

b) 0

c) 5.948

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.16 ounces

Standard Deviation, σ = 0.08 ounces

We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) Standard deviation of 23 bags

$\displaystyle\frac{S.D}{\sqrt{23}} = \frac{0.08}{\sqrt{23}} = 0.0167$

b) P( fill volume of 23 bags is below 5.95 ounces)

P(x < 5.95)

$P( x < 5.96) = P( z < \displaystyle\frac{5.95 - 6.16}{0.0167}) = P(z < -12.57)$

$= 1 - P(z < 12.57)$

Calculation the value from standard normal z table, we have,

$P(x < 5.95) = 1 - 1 = 0$

c) P( fill volume of 23 bags is below 6 ounces) = 0.001

P(x < 6) = 0.001

$P( x < 6) = P( z < \displaystyle\frac{6 - \mu}{0.0167})$

Calculation the value from standard normal z table, we have,

$P( z \leq -3.09) = 0.001$

$\displaystyle\frac{6 - \mu}{0.0167} = -3.09\\\\\mu = 6 + (0.0167\times -3.09) = 5.948$

If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.

7 0

3 years ago