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3 years ago

You have a map of Southern California. If the distance between

Mathematics

1 answer:

Rudiy273 years ago

4 0

9514 1404 393

Answer:

18 miles

Step-by-step explanation:

The map distance and ground distance are proportional.

(distance to school)/(3 in) = (12 miles)/(2 in)

Multiplying by 3 in gives ...

distance to school = (12 miles)(3/2) = 18 miles

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Irina-Kira [14]

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Step-by-step explanation:

$\frac{x+ 3}{12x} \ \cdot \ \frac{4x}{x^2 +x - 6}\\\\= \frac{x+ 3}{12x} \ \cdot\ \frac{4x}{x^2 +3x - 2x - 6}\\\\= \frac{x+ 3}{12x} \cdot \frac{4x}{x(x + 3) -2(x+3)}\\\\= \frac{x+ 3}{12x} \cdot \frac{4x}{(x + 3)(x -2)}\\\\= \frac{1}{12x} \ \cdot \ \frac{4x}{(x - 2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\\= \frac{1}{3} \ \cdot \ \frac{1}{(x - 2)}\\\\= \frac{1}{3(x - 2) }$

5 0

3 years ago

I need help with this question i dont understand it

RUDIKE [14]

The perimeter, by definition, is the outside measure of that figure. MN and LM are the same length and LK and NK are the same length....we just need to find the lengths! Use the distance formula to find the distance between the 2 points:
$\sqrt{( x_{2} - x_{1} ) ^{2} +( y_{2}- y_{1} ) ^{2} }$
For the segment MN, use the coordinates of M as your x1, y1, and use the coordinates of N for x2, y2:
$\sqrt{(3-2) ^{2}+(4-3) ^{2} }$
which simplifies to
$\sqrt{(1 )^{2}+(1) ^{2} }$ which is $\sqrt{2}$
So that is the length of both MN and LM. So far our perimeter is $\sqrt{2} + \sqrt{2}=2 \sqrt{2}$
Now let's use the same formula to find out the length of one of the longer segments:
$\sqrt{(5-3) ^{2} +(3-2) ^{2} }$
which simplifies down to
$\sqrt{(2) ^{2} +(1) ^{2} }$
which is of course $\sqrt{5}$
Since we have 2 of those lengths, $\sqrt{5} + \sqrt{5}=2 \sqrt{5}$
So our perimeter is, in the end, $2 \sqrt{2}+2 \sqrt{5}$
That's the third choice down

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2 years ago