Let g(x) be the indicated transformation of f(x)=3x+1. vertical stretch by a factor of 2

Nance is riding her bike to a friend’s house 4 and 1/2 miles away. She’s ridden 2 and 3/5 miles. How much farther does she need

Alinara [238K]

<h2>Hello!</h2>

The answer is:

She still needs to ride

$1\frac{9}{10}miles$

From the statement, we know that Nance's friend is 4 1/2 miles away and she only has ridden 2 3/5 miles.

So, let be "x" the distance that she still need to ride to get to her friend's house.

The equation will be:

$TotalDistance=RiddenDistance+x$

Remember that:

$a\frac{b}{c}=a+\frac{b}{c}$

Also, if we need to add/substract fractions we must remember that:

$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$

If we need to switch fractions to mixed numbers, we need to:

- Perform the division

- Keep the whole number and make a fraction with the aproximated next whole number (remainder) as numerator, and the original denominator. We need to use the remainder to approximate to the next whole number.

$\frac{a}{b}=c.r$

Where:

$a=numerator\\b=denominator\\c=WholePart\\r=NextWholeNumber$

So,

$\frac{a}{b}=c\frac{r}{b}$

Then, substituting the given information into the equation, we have:

$4\frac{1}{2} =2\frac{3}{5} +x\\\\x=(4\frac{1}{2}) -(2\frac{3}{5})=4+\frac{1}{2}-2-\frac{3}{5}=2+\frac{1}{2}-\frac{3}{5}\\\\x=2+\frac{1}{2}-\frac{3}{5}=2+\frac{5-6}{10}=2-\frac{1}{10}\\\\x=2-\frac{1}{10}=\frac{20-1}{10}=\frac{19}{10}$

So, she still need to ride $\frac{19}{10}miles$

or

$1\frac{9}{10}miles$

To prove if the answer is reasoable, let's substituite the obtained value into the equation and check if the equation is satisfied:

$4\frac{1}{2}=2\frac{3}{5}+1\frac{9}{10}$

$4\frac{1}{2}=\frac{13}{5}+\frac{19}{10}$

$4\frac{1}{2}=\frac{9}{2}$

$4\frac{1}{2}=4\frac{1}{2}$

Have a nice day!

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To solve this system of equations by elimination, what operation could be used to eliminate the x-variable and find the value of

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The system of equations are:

2x − 4y = 6
−3x + 3y = 9

To eliminate x, you should:
1. multiply first equation with 3
2. multiply second equation with 2
3. add the new equations you got

So your answer is A.

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