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Dmitriy789 [7]
3 years ago
14

If the point (x, y) is in Quadrant III, which of the following must be true?

Mathematics
1 answer:
strojnjashka [21]3 years ago
3 0
X must be greater than 0 and y must be less than 0.
I hope this helped.
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In two or more complete sentences write and solve an equation for the situation and explain how you will solve the equation. Fif
miv72 [106K]

We conclude that the starting average grade is 79.17%

<h3>How to write the equation and solve it?</h3>

There are 50 students, let's say that the grades are measured between 1% and 100%.

And the average grade of the 50 students is A.

We know that after it increased by 20%, the average of the grades is 95.

Then we just need to solve the percentage equation:

95 = A*(1 + 20%/100%) = A*(1.2)

95/1.2 = A = 79.17

We conclude that the starting average grade is 79.17%

If you want to learn more about percentages:

brainly.com/question/843074

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6 0
2 years ago
Find the area of the parallelogram by composing into rectangles or decomposing into triangles or other shapes as needed. 8 yd 6
Gennadij [26K]

Answer:

Below.

Step-by-step explanation:

The area = area of a rectangle on the same base and between the same parallel lines

= (15 + 6) * 8

= 21 * 8

= 168 yd^2.

8 0
3 years ago
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two teams are playing in the finals for an adult soft leauge. Each team has 13 players and the ages of the team members are show
Andrew [12]

Answer: C.

Step-by-step explanation:

6 0
3 years ago
Which of the following is not true?
cestrela7 [59]

The last statement is NOT true.

1/2 divided by 4 is 0.125 and 1/2 times 4/1 is 2

Hope I helped!

4 0
3 years ago
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PLEASE HELP it should be easy for someone but I can’t do it
-Dominant- [34]

Answer:

Total crackers on the plate are 12

Step-by-step explanation:

Manuel ate crackers = \frac{1}{3}

His brother ate crackers = \frac{1}{4}

Crackers left on the plate = 5

We need to find how many crackers were there on the plate.

Let x be the total crackers on the plate

So, we can write the equation

x-\frac{1}{3}x-\frac{1}{4}x=5

Because 1/3 and 1/4 crackers are eaten and 5 are left so, we subtract 1/3x and 1/4x from x and equal it to 5

\frac{12x-4x-3x}{12}=5\\\frac{12x-7x}{12}=5\\\frac{5x}{12}=5\\x=\frac{5*12}{5}\\x=12

So, we get x = 12

Therefore, total crackers on the plate are 12

8 0
3 years ago
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