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alexandr402 [8]
3 years ago
12

The diagram below represents a relation. List the ordered pairs in the relation and tell whether or not the relation is also a f

unction.
Mathematics
1 answer:
cricket20 [7]3 years ago
3 0

A relation is a set of inputs and outputs, often written as ordered pairs (input, output). We can also represent a relation as a mapping diagram or a graph. For example, the relation can be represented as:

 
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Ellen drove 357.9 miles. Her car gets about 21 miles per gallon. Which is the best estimate of how many gallons of gas Ellen use
olga55 [171]
The best thing to do is to divide the amount she has driven (357.9) by the miles (21)
As you're looking for an estimate, you have to round them.
357.9 rounds to 360
21 rounds to 20
Therefore, you've got to divide them
360/20= 18
Therefore, Ellen has driven approximately 18 miles
Hope this helps :)
7 0
3 years ago
Read 2 more answers
5/6w+21=-1/3(2w-9).
oksian1 [2.3K]
I believe this question logically tells us to find the value of w. The two equations are already equated. Since there is 1 unknown and 1 equation, the system is solvable. The solution is as follows:

5/(6w+21) = -1/3(2w - 9)
5/(6w+21) = -1/(6w - 27)
Cross multiplying the terms:
5(6w - 27) = -1(6w +21)
30w - 135 = -6w - 21
30w + 6w = -21 + 135 = 114
36w = 114
w = 114/36
w = 19/6 or 3.167
7 0
3 years ago
Pls helpppppp this stuff makes no sense
yulyashka [42]

Answer:

x = 5

Step-by-step explanation:

a || b and a line intersecting them is their transversal.

\therefore m\angle 1=m\angle 5\\(corresponding \:\angle 's) \\\\\therefore 60 - 2x = 70 - 4x\\\\\therefore 4x - 2x = 70 - 60\\\\\therefore 2x = 10\\\\\therefore x =\frac{10}{2} \\\\\huge \orange {\boxed {\therefore x = 5}}

4 0
3 years ago
Plz help me out a bit :(<br> which one do you think it is?
Anna [14]
C. 9^4
when dividing with exponents, i usually just look at the exponents and subract them so i did 12-8 for this problem
8 0
2 years ago
Read 2 more answers
Let f(x)=x^3 and g(x)= √x, and h(x)=x/3. Find each of the following: f(g(h(6)))
rewona [7]

Answer:

go from inside out.

h(x)=x/3

(\sqrt{2})^{3}=> h(6) = 6/3 = 2

g(x) = \sqrt{x}

=> g(h(6)) = g(2) = \sqrt{2}

f(x) = x^{3}

=> f(g(h(6))) = f(\sqrt{2} ) = (\sqrt{2} )^{3}

6 0
3 years ago
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