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omeli [17]
3 years ago
13

Jalil is training for a marathon. On Monday, he biked 7.2 miles from his home to a park where he ran 10 laps on a path around a

small pond. Then he biked home. Jalil biked and ran a total of 16.9 miles. What is the distance of each lap around the pond?
Please answer quickly.
Mathematics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

.25 or 1/4 mile

Step-by-step explanation:

7.2 miles biking x 2 (round trip)= 14.4 miles

Total miles is 16.9-14.4= 2.5 miles left to run 10 times around a small pond. So to find out the distance of each lap around the pool you divide 2.5 miles by 10 laps equals .25 or 1/4 Mike is the distance around the pond.

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.. Which of the following are the coordinates of the vertices of the following square with sides of length a?
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Option A: O(0,0), S(0,a), T(a,a), W(a,0)

Option D: O(0,0), S(a,0), T(a,a), W(0,a)

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Option A: O(0,0), S(0,a), T(a,a), W(a,0)

To find the sides of a square, let us use the distance formula,

d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Now, we shall find the length of the square,

\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}

Thus, the square with vertices O(0,0), S(0,a), T(a,a), W(a,0) has sides of length a.

Option B: O(0,0), S(0,a), T(2a,2a), W(a,0)

Now, we shall find the length of the square,

\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}

This is not a square because the lengths are not equal.

Option C: O(0,0), S(0,2a), T(2a,2a), W(2a,0)

Now, we shall find the length of the square,

\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}

Thus, the square with vertices O(0,0), S(0,2a), T(2a,2a), W(2a,0) has sides of length 2a.

Option D: O(0,0), S(a,0), T(a,a), W(0,a)

Now, we shall find the length of the square,

\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}

Thus, the square with vertices O(0,0), S(a,0), T(a,a), W(0,a) has sides of length a.

Thus, the correct answers are option a and option d.

8 0
3 years ago
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