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Bas_tet [7]
3 years ago
11

I need the work for #31 & #37

Mathematics
1 answer:
Aleks [24]3 years ago
7 0
Problem 31)

The terms 10s and -8s are like terms. This is because they have the same variable term: s. 

What we do is simply subtract the numbers in front of the letter s like so 10-8 = 2

So 10s - 8s = 2s

Think of it like this: we have 10 sodas and we take away 8 sodas. We'll be left with 2 sodas. Writing 10s-8s = 2s is a way to use shorthand. 

So overall, the original expression 5 + 10s - 8s turns into 5 + 2s which is the final answer

Note: 5 + 2s can be written as 2s + 5. The order in which we add things up doesn't matter.

-----------------------------------------------------------------------------------

Problem 37)

In this problem, the terms 5 and -7 are like terms. They are the terms without any variables (letters) attached to them. 

If we have 5 dollars and we want to spend 7 dollars to buy something, then we'll be 2 dollars in debt. Use negative numbers to indicate debt. 

So 5 - 7 = -2

The expression 5 + (4g - 7) simplifies to 4g - 2 which is the final answer
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Answer:

34.26miles from 49.58miles of the drive (69.11% of the drive).

Step-by-step explanation:

A problem of Analytic Geometry. This question can be solved using  the resulting values from equaling the equations of the line (for the drive) and circle (with radius equals 30miles for radio transmitter broadcasting), and calculating the total distances from coincident points between circle and line, and total drive.

A graph is attached showing part of the circle and line with coincident points.

<h3>Line's Equation</h3>

Assuming transmitter is located (0, 0). From a city 33 miles north of the transmitter (0, 33) to a second city 37 miles east of the transmitter (37, 0).

Then, the equation for the line is (taking the two points from the start to the end):

\\ y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}*(x-x_{1})

\\ y-33=\frac{0-33}{37-0}*(x-0)}

\\ y-33=\frac{-33}{37}*x}

\\ y=\frac{-33}{37}*x} + 33 [1]

<h3>Circle's Equation</h3>

The equation for the circle whose center (0, 0) is:

\\ (x-0)^2 + (y-0)^2 = 30^2

\\ x^2 + y^2 = 30^2

\\ y^2 = 30^2 - x^2

\\ y = \frac{+}{-}\sqrt{30^2- x^2} [2]

Equaling [1] and [2], to determine the points where starting to receive the radio transmitter signals to the point where finishing those signals:

\\ \frac{-33}{37}*x + 33 = \sqrt{30^2- x^2}

Solving this equation for <em>x</em>, we have two solutions for it (from <em>WolframAlpha</em>):

\\ x_{1} = 40293/2458 - (111 \sqrt(80151))/2458 \approx 3.60775

\\ x_{2} = 40293/2458 + (111 \sqrt(80151))/2458 \approx 29.1774

Then, using [1], the corresponding values for <em>y</em> are:

\\ y_{1}=\frac{-33}{37}*(3.60775) \approx 29.78

\\ y_{2}=\frac{-33}{37}*(29.1774) \approx 6.97

So,

\\ (x_{1}=3.61, y_{1}=29.78)

\\ (x_{2}=29.18, y_{2}=6.97)

Well, the distance of the drive is:

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

\\ d = \sqrt{(29.18-3.61)^2 + (6.97-29.78)^2}

\\ d \approx 34.2654 miles.

The total distance traveled is:

\\ d = \sqrt{(37-0)^2 + (0-33)^2}

\\ d \approx 49.5782 miles

Thus, during the drive, the signal of the radio transmitter was picked up for 34.26miles from the total of the 49.58 traveled, that is, a fraction equivalent to \\ \frac{34.2654}{49.5782} \approx 0.6911 or 69.11% of the drive.

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