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Nastasia [14]
3 years ago
8

Lenny is competing with his cousin, Jasper, in an indoor rock-climbing contest. At the start of the climb, Lenny makes his way 5

¼ feet up the wall, while Jasper climbs 9 ¾ feet. How much farther did Jasper climb than Lenny?
Mathematics
1 answer:
SVEN [57.7K]3 years ago
4 0

Answer:

4\frac{1}{2} feet further.

Step-by-step explanation:

Since these are mixed numbers that are both in fourths, we can easily subtract the two numbers. However, I find it easier if we first convert both mixed numbers into improper fractions.

5\frac{1}{4} = \frac{5\cdot4+1}{4}  = \frac{21}{4}

9\frac{3}{4} = \frac{9\cdot4+3}{4}  = \frac{39}{4}

Now we can subtract the numerators:

\frac{39}{4} - \frac{21}{4} = \frac{39-21}{4} = \frac{18}{4}

\frac{18}{4} simplifies down to \frac{9}{2}.

Converting  \frac{9}{2} to a mixed number is easy - 2 goes into 9 4 times (8) with one remainder so:

4\frac{1}{2} .

Hope this helped!

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Yes; Opposite sides are congruent, and diagonals are congruent.

Step-by-step explanation:

we have

A(4, -7), B(4, -2), C(0, -2), D(0, -7)

we know that

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step 1

Find the length of the sides

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substitute the values

d=\sqrt{(-2+7)^{2}+(4-4)^{2}}

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AB=5\ units

<u><em>Find the distance BC</em></u>

substitute the values

d=\sqrt{(-2+2)^{2}+(0-4)^{2}}

d=\sqrt{(0)^{2}+(-4)^{2}}

BC=4\ units

<u><em>Find the distance CD</em></u>

substitute the values

d=\sqrt{(-7+2)^{2}+(0-0)^{2}}

d=\sqrt{(-5)^{2}+(0)^{2}}

CD=5\ units

<u><em>Find the distance AD</em></u>

substitute the values

d=\sqrt{(-7+7)^{2}+(0-4)^{2}}

d=\sqrt{(0)^{2}+(-4)^{2}}

AD=4\ units

Compare the length sides

AB=CD

BC=AD

therefore

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step 2

Find the length of the diagonals

<u><em>Find the distance AC</em></u>

substitute the values

d=\sqrt{(-2+7)^{2}+(0-4)^{2}}

d=\sqrt{(5)^{2}+(-4)^{2}}

AC=\sqrt{41}\ units

<u><em>Find the distance BD</em></u>

substitute the values

d=\sqrt{(-7+2)^{2}+(0-4)^{2}}

d=\sqrt{(-5)^{2}+(-4)^{2}}

BD=\sqrt{41}\ units

Compare the length of the diagonals

AC=BD

therefore

Diagonals are congruent

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