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Zinaida [17]
3 years ago
12

Please help and explain!

Mathematics
1 answer:
Scrat [10]3 years ago
5 0
K is the slope of the line formed by connecting the points.
Use slope formula:
\frac{y_2 - y_1}{x_2 - x_1}
where the 2 points are the end points of the line (First and last going Left to right)
(x_1,y_1) = (2,7)  \\ (x_2,y_2) = (8,28)
Substitute into slope formula
k= \frac{28-7}{8-2} = \frac{21}{6} = \frac{7}{2}
You can check if this is correct by going back to graph and going "up 7" and "over 2" to get from one point to the next.
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Find the area and perimeter of trapezium.
neonofarm [45]

Answer:

Perimeter- 26cm

Area- I Don't know sorry. I'll comment if I figure it out

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2 years ago
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Help please, and explain.
Korvikt [17]

Answer: 2x + y

<u>Step-by-step explanation:</u>

logₐ(3) = x

logₐ(5) = y

logₐ(45) = logₐ(3²· 5)

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             = 2 logₐ(3) + logₐ(5)

             = 2     x      +   y        <em>substituted given values (stated above)</em>


4 0
3 years ago
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When a number is added to 1/5 of itself, the result is 24. The equation that models this problem is n + 1/5n = 24. What is the v
Zielflug [23.3K]
The answer is 20 since 1/5 of 20 is 4, and 20 + 4 = 24.
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3 years ago
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An octagonal swimming pool has a base area of 22 square feet The pool is 3 feet deep How many cubic feet of water can the pool h
LenKa [72]

The pool can hold 65.84 ft³ of water

<u>Explanation:</u>

Given:

Shape of pool = octagonal

Base area of the pool = 22 ft²

Depth of the pool = 3 feet

Volume, V = ?

We know:

Area of octagon = 2 ( 1 + √2) a²

22 ft² = 2 ( 1 + √2 ) a²

\frac{11}{1+\sqrt{2} } = a^2

a² = \frac{11}{2.42}

a² = 4.55

a = 2.132 ft

Side length of the octagon is 2.132 ft

We know:

Volume of octagon = 2(1+\sqrt{2} ) X (a)^2 X h

V = 2(1+\sqrt{2})X (2.132)^2 X 3\\ \\V = 2 ( 2.414) X 4.5454 X 3\\\\V = 65.84 ft^3

Therefore, the pool can hold 65.84 ft³ of water

8 0
3 years ago
Don't know need help [pls
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Answer:

56

......................................................

8 0
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