What is the slope of the line that represents this relationship?

Mathematics

1 answer:

galben [10]3 years ago

7 0

Answer:

undefine

Step-by-step explanation:

the slope does not show rise over run it just sgow one of them therefore it is undefine

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What is 2/9 + 1/9 + 1/6 in fraction form

Nikitich [7]

We get all these to have the same denominators for all the fractions.

2*2/9*2 = 4/18
1*2/9*2 = 2/18
1*3/6*3 = 3/18

Now we can add all the numerators.

4/18 + 2/18 + 3/18 = 9/18

Simplify our answer by dividing both numbers by 3, then by 2.

9/3 and 18/3 = 3/6= 1/2

Our final answer is 1/2

6 0

3 years ago

Julian walked 6/10 of a mile to his friends house and another 35/100 mile to the store. He walked 1/4 of a mile back home. Julia

AveGali [126]

The statement, "Julian's sister said he walked 1/5 mile" cannot be agreed because Julian totally walked $1\frac{1}{5} \text{ or } \frac{6}{5}$ miles.

<u>Solution:</u>

Given that,

Julian walked 6/10 of a mile to his friends house
Another 35/100 mile to the store
He walked 1/4 of a mile back home

To find total distance walked by Julian we have to add the above stated values. That is, $\frac{6}{10} +\frac{35}{100} +\frac{1}{4}$

Factors of 10 = $5\times2$

Factors of 100 = $5\times2\times5\times2$

Factors of 4 = $2\times2$

Therefore, the least common factor of 10, 100 and 4 is 100. With like denominators we can operate on just the numerators,

$\frac{6\times10}{10\times10} +\frac{35\times1}{100\times1} +\frac{1\times25}{4\times1}\rightarrow\frac{60+35+25}{100}\rightarrow\frac{120}{100}$

$\Rightarrow\frac{120}{100}\rightarrow\frac{6}{5}$

Which can also be written as $1\frac{1}{5}$ .

So, from the above calculation it can be said that Julian walked $1\frac{1}{5} \text{ miles }$ .

3 0

3 years ago

In the following table of values, the first differences are

ExtremeBDS [4]

Answer:

-7

Step-by-step explanation:

Given

$\begin{array}{ccc}x & {y} & {First\ difference} & {-2} & {34} & {}&{-1} & {27} & {} &{0} & {20} & {} & {1} & {13} & {} & {2} & {6} &{} \ \end{array}$