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3 years ago

FIND THE DISTANCE BETWEEN THE TWO POINTS A(0, 0), B(6, 8)

Mathematics

1 answer:

Alex3 years ago

7 0

Answer:

Step-by-step explanation:

Mathematical Formula: $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Therefore: $\sqrt{\left(6-0\right)^2+\left(8-0\right)^2}$

Which = 10

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Simplify (Remember to simplify means to write an equivalent expression that uses each variable on')

DaniilM [7]

Answer:

A) $x^{3} y^{4}$

B) $-\frac{5x^{2} }{y}$

C) $9m^{4}n^{4}$

D) $\frac{1}{x^{3} }$

Step-by-step explanation:

Hope this helps.

6 0

2 years ago

Complete the table. Round to the nearest cent if necessary.

murzikaleks [220]

Answer: 99.07$
Steps: First turn 32% into a decimal by moving the decimal right 2 places. You get 0.32. Then multiply 75.05$ by your decimal. You get 24.016. Add that to 75.05 and round your cent up, getting 99.07$ Sorry if i am incorrect this is my first answer. Hope I helped :)

6 0

3 years ago

dylan started working on homework at 5:45 p.m. it took him 1 hour and 57 minutes to complete it what time did

kkurt [141]

I'm assuming its asking what time he finished? I added it up and it seems to be 7:42 pm.

5 0

3 years ago

Write the equation of the line that passes through the points (4,-6) and (-2,3)

yarga [219]

Here is your answer and an explanation

Answer: y = -2/3x

Explanation:

This can be determined by calculating the gradient of the straight line, using:

m=ΔyΔx

=−6−34−(−2)

=−96

=−32

Then we use the slope-point form of the straight line:

y−y1=m(x−x1)

to give:

y−3=−32(x−(−2))

∴y−3=−32(x+2)

∴y−3=−32x−3

∴y=−32x

7 0

3 years ago

PLZ HELP !! Which of the points (s, t) is not one of the vertices of the shaded region of the

stira [4]

Answer:

A(0,10)

Step-by-step explanation:

Given the 4 inequalities:

s ≥ 12 - 0.5t (1)

s ≥ 10 -t (2)

s ≤ 20-t (3)

s ≥ 0

t ≥ 0

Let analyse all 4 possible answer:

A(0, 10)

Let substitute this point into (1) we have: 10 ≥ 12 -0.5*0 Wrong

We do not choose A

B (0, 20)

Let substitute this point into (1) we have: 20 ≥ 12 -0.5*0 True

Let substitute this point into (2) we have: 20 ≥ 10 - 0 True

Let substitute this point into (3) we have: 20 ≤ 20 - 0 True

We choose B as the vertex

C. (4, 16)

Let substitute this point into (1) we have: 16 ≥ 12 -0.5*4 True

Let substitute this point into (2) we have: 16 ≥ 10 - 4 True

Let substitute this point into (3) we have: 16 ≤ 20 - 4 True

We choose C as the vertex

D. (16,4)

Let substitute this point into (1) we have: 4 ≥ 12 -0.5*16 True

Let substitute this point into (2) we have: 4 ≥ 10 - 16 True

Let substitute this point into (3) we have: 4 ≤ 20 - 16 True

We choose B as the vertex

Hence, the point A(0,10) is not one of the vertices of the shaded region of the set of inequalities.

3 0

3 years ago