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stepladder [879]
2 years ago
15

State the domain and range and determine if it's a function?

Mathematics
1 answer:
pishuonlain [190]2 years ago
4 0

if we do a <u>vertical line test</u> on that graph, we'll find that a vertical line will pass by and hit the line only once on its way down, only once meaning the graph is the graph of a function.

<h2>Range</h2>

well, is really how high and low it goes or namely over the y-axis.

well, it goes up up up to 3 then U-turns and back down it goes, now the graph has arrowheads, meaning the graph keeps on going towards infinity, vertically as well as horizontally since it's a parabola, so the range will be

<h2>[3 , +∞)</h2>

<h2>Domain</h2>

well, is simply how left and right it goes or namely over the x-axis.

judging from the arrowheads it moves from infinity to infinity, so

<h2>(-∞ , +∞).</h2>
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PLEASE HELP!!!
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Answer/Step-by-step explanation:

Part A: Net A is the correct net. If we decide to fold the net, we'd get the shape of the prism. Folding back net B won't give us the shape of the prism because of the position and arrangement of the right triangular  bases.

Part B:

AB =  3in

BC = 5 in

CD = 9.4 in

Part C: surface area of the prism can be calculated by calculating the area of each part of the net, and summing them together as follows,

Area of the 2 triangular bases = 2(½*base*height of triangle) = 2(½*4*3) = 2*2*3 = 12 in²

Area of the three rectangles =

= (9.4*4) + (9.4*3) + (9.4*5) = 37.6 + 28.2 + 47 = 112.8 in^2

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2 years ago
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the answers that are equivalent are a, b, and e

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2 years ago
A motorboat is capable of traveling at a speed of 14 miles per hour in still water. On a particular day, it took 15 minutes long
Anon25 [30]

By solving a system of equations we will find that the rate of current in the stream is S = 2 mi/h.

When the motorboat travels downstream, the total velocity will be the velocity of the motorboat in still water plus the velocity of the stream, while if the motorboat travels upstream, we have the velocity of the stream subtracted.

So upstream the speed is:

(14 mi/h - S)

Downstream the speed is:

(14 mi/h + S)

Where S is the rate of current in the stream.

We know that downstream it takes 15 minutes more to travel 12 miles, then we can write the system of equations:

(14 mi/h + S)*T = 12 mi

(14 mi/h - S)*(T - 15 min) = 12 mi

To solve this, we need to isolate one of the variables in one of the equations, I will isolate T in the first one:

T = (12 mi)/(14 mi/h + S)

Replacing that in the other equation we will get:

(14 mi/h - S)*((12 mi)/(14 mi/h + S) - 15 min) = 12 mi

Now we can solve this for S. Now we can multiply both sides by (14 mi/h + S).

(14 mi/h - S)*12 mi  - (14 mi/h + S)*(14 mi/h - S)*(- 15 min) = 12 mi*(14 mi/h + S)

Also notice that the speeds are in hours, so we can rewrite:

- 15 min = -0.25 h

(14 mi/h - S)*12 mi  - (14 mi/h + S)*(14 mi/h - S)*(- 0.25 h) = 12 mi*(14 mi/h + S)

168 mi^2/h - 12mi*S  + 49mi^2/h + 0.25h*S^2 = 168mi^2/h + 12mi*S

- 12mi*S  + 49mi^2/h - 0.25h*S^2 = 12mi*S

-24mi*S -  0.25h*S^2  + 49mi^2/h = 0

This is a quadratic equation, the solutions are:

S = \frac{24mi \pm \sqrt{(-24mi)^2 - 4*(49mi^2/h)*(-0.25h)}  }{2*-0.25h} \\\\S =  \frac{24mi \pm 25 mi  }{-0.5h}

We only take the positive solution, so we get:

S = (24 mi - 25 mi)/(-0.5 mi) = 2 mi/h

The rate of current in the stream is 2 mi/h.

If you want to learn more, you can read:

4 0
2 years ago
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