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LUCKY_DIMON [66]
3 years ago
15

erry jogs the same round trip path everyday. He jogs 0.7 miles to the park and then 0.5 miles from the park to the playground. A

fter getting a drink at the playground, he then jogs back from the playground to his house. He jogs from the playground to the bridge 0.3 miles and then 0.9 miles from the bridge to his house. When Jerry completes his jog, how many miles is he from his house?
Mathematics
2 answers:
Reil [10]3 years ago
7 0

Answer:

0 miles

Step-by-step explanation:

adelina 88 [10]3 years ago
4 0

To solve this problem you must use the information given in the problem above:

1. You have that he jogs 0.7miles to the park and then 0.5miles to the playground, then he jogs 0.3 miles to the bridge and 0.9miles to his house.

2. Therefore, as you can see, Jerry completes his jog when he arrives to his house. So, he is 0miles from his house.

Then, the answer is: 0miles

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Basile [38]

We are given a right triangle that has angles of 45°-45°-90°. This would indicate that the triangle is an isosceles type. We can use some trigonemetric functions to solve for the other legs. We do as follows:

 

sin 45 = p / 10

p = 5√2

 

cos 45 = q /10

q = 5√2

 

<span>Hope this answers the question. Have a nice day.</span>

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
Plz help<br>urgent !!!!<br>will give the brainliest !!​
Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

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Step-by-step explanation:

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