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Alchen [17]
2 years ago
11

Use the following graph to find the slope of the line connecting Saturn and earth

Mathematics
2 answers:
marissa [1.9K]2 years ago
5 0

Answer:

The slope is 2.

Step-by-step explanation:

To find the slope we can take two points from the graph (in this example I'll use (3, 2) and (4, 4)) and input them in this expression: \frac{y2-y1}{x2-x1}

\frac{4-2}{4-3}

Solve:

4 - 2 = 2

4 - 3  = 1

\frac{2}{1} = 2

The slope of the line connecting Saturn and Earth is 2.

Hope this helps :)

anyanavicka [17]2 years ago
3 0
2 is the correct slope:)
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A cafe sells puffs. 5/9 are beef and 1/3 are chicken. The remaining 18 are tuna. Calculate 1. What fraction of the puffs are tun
Ksivusya [100]
Let n be the total number of puffs. Now we can write:
n=\frac{5}{9}n+\frac{1}{3}n+18
Adding the fractional value of n, we get:
n=\frac{5+3}{9}n+18
Simplifying and rearranging gives us:
n-\frac{8}{9}n=18
Therefore we can simplify to get:
\frac{1}{9}n=18
and finally
n=18\times9=?
The fraction of the puff that are tuna is found from:
\frac{18}{18\times9}=\frac{1}{9}
7 0
3 years ago
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35 into binary number<br>​
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Answer:

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Step-by-step explanation:

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What is 7285 divided by 4
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3 years ago
7x-11=5(x-2)+2x-1= ?​
yan [13]
0 = 0


Simplifying
7x + -11 = 5(x + -2) + 2x + -1

Reorder the terms:
-11 + 7x = 5(x + -2) + 2x + -1

Reorder the terms:
-11 + 7x = 5(-2 + x) + 2x + -1
-11 + 7x = (-2 * 5 + x * 5) + 2x + -1
-11 + 7x = (-10 + 5x) + 2x + -1

Reorder the terms:
-11 + 7x = -10 + -1 + 5x + 2x

Combine like terms: -10 + -1 = -11
-11 + 7x = -11 + 5x + 2x

Combine like terms: 5x + 2x = 7x
-11 + 7x = -11 + 7x

Add '11' to each side of the equation.
-11 + 11 + 7x = -11 + 11 + 7x

Combine like terms: -11 + 11 = 0
0 + 7x = -11 + 11 + 7x
7x = -11 + 11 + 7x

Combine like terms: -11 + 11 = 0
7x = 0 + 7x
7x = 7x

Add '-7x' to each side of the equation.
7x + -7x = 7x + -7x

Combine like terms: 7x + -7x = 0
0 = 7x + -7x

Combine like terms: 7x + -7x = 0
0 = 0

Solving
0 = 0

3 0
3 years ago
If 270°&lt;θ&lt;360°, and tan⁡ θ= - 1/2, what is the value of tan⁡(−θ)?
stiv31 [10]

Check the picture below.

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since we know that θ is on the IV Quadrant, then its negative counterpart must be across the line, and tangent is negative on the IV Quadrant, and positive on the I Quadrant.

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