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mixas84 [53]
3 years ago
14

Please answer! Explain all steps!

Mathematics
2 answers:
MatroZZZ [7]3 years ago
7 0
We will be using the formulas:
speed=distance/time
time=distance/speed
distance=speed×time
First let's find out Diane's rate of swimming. We can measure this by finding the slope (y/x) of a given coordinate on the graph. One point is (10,15), so you do 15/10=1.5m/s
Now for Rick's rate of swimming, just take a pair of values from the table. 12.5/10=1.25m/s
By the way m/s is metres per second for this
So at a constant speed of 1.5m/s, Diane swam 150m in 150/1.5= 100 seconds, or 1 minute 40 seconds
And at a constant speed of 1.25m/s, Rick swam 150m in 150/1.25= 120 seconds, or 2 minutes.
So the difference between their two times is 20 seconds
Harlamova29_29 [7]3 years ago
3 0
From the grid on Diane's swim you can see that, for every two squares on the grid over the x-axis, it goes up 3 squares over the y-axis, it moves 2 to the right and then 3 up, and you get the next point.  What does that mean?  well, is a constant speed and thus the graph is a line, with a slope of 3 meters per 2 seconds, so her slope is 3/2 m/s.

now, for Rick's slope, we can just pick two points off of it, say, hmmm 10, 12.5 and 20, 25, and get the slope,

 \bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~{{ 10}} &,&{{ 12.5}}~) 
%  (c,d)
&&(~{{ 20}} &,&{{ 25}}~)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}\implies 
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{25-12.5}{20-10}\implies \cfrac{12.5}{10}\implies \cfrac{\frac{125}{10}}{10}
\\\\\\
\cfrac{\frac{125}{10}}{\frac{10}{1}}\implies \cfrac{125}{10}\cdot \cfrac{1}{10}\implies \cfrac{125}{100}\implies \cfrac{5}{4}\cdot \cfrac{meters}{second}

so Diane is doing 3 meters for every 2 seconds, and Rick is doing 5 meters for every 4 seconds.

how long will it be for each to do the 150 meters anyway?

\bf ~~~~~~~~~~~~~~Diane's\\\\
\begin{array}{ccll}
meters&seconds\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
3&2\\
150&x
\end{array}\implies \cfrac{3}{150}=\cfrac{2}{x}\implies x=\cfrac{150\cdot 2}{3}\implies x=100\\\\
-------------------------------\\\\

\bf ~~~~~~~~~~~~~~Rick's\\\\
\begin{array}{ccll}
meters&seconds\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
5&4\\
150&y
\end{array}\implies\cfrac{5}{150}=\cfrac{4}{y}\implies y=\cfrac{150\cdot 4}{5}\implies y=120

what's their difference?  well y - x.
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Hey there! I'm happy to help!

The domain is any possible number you can input into the function to get a real output. The domain of h just means the domain of this entire function, which is called h.

Let's look at the answer options.

OPTION A

All real values of x such that x≠0.

The only way to make it so that we do not have a real output is if we get a negative square root. You cannot multiply any number by itself to get a negative number unless you use imaginary numbers, but using imaginary numbers makes our output not real.

Anyways, plugging in 0 would give us √-10, which is not a real number. That part is correct, but this option says ALL REAL NUMBERS except for 0. The problem is is that we can take any number less than ten and plug it in and we would get a negative square root, a fake number. So, this option is incorrect.

OPTION B

All real values of x such that x≥10.

Let's say we use 10 for our x and plug it in. This gives us √0, which is 0, a real output. Anything bigger than this 10 will give us a real output as well, so this option is correct.

We don't even need to check the other options because we have already found the correct answer. C,D, and E are all incorrect though because they include values less ten, which would give us a negative square root, a fake number.

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3 years ago
Prove ΔPAB is isosceles.
Licemer1 [7]

Answer:

See explanation

Step-by-step explanation:

If PX\cong PY, then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so

\angle 1\cong \angle 2

and

m\angle 1=m\angle 2

Angles 1 and 3 are supplementary, so

m\angle 3=180^{\circ}-m\angle 1

Angles 2 and 4 are supplementary, so

m\angle 4=180^{\circ}-m\angle 2

By substitution property,

m\angle 4=180^{\circ}-m\angle 2=180^{\circ}-m\angle 1=m\angle 3

Hence,

\angle 3\cong \angle 4

Consider triangles APX and BPY. In these triangles:

  • PX\cong PY - given;
  • \angle 5\cong \angle 6 - given;
  • \angle 3\cong \angle 4 - proven,

so \triangle APX\cong \triangle BPY by ASA postulate.

Congruent triangles have congruent corresponding sides, then

AP\cong BP

Therefore, triangle APB is isosceles triangle (by definition).

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Answer:

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Step-by-step explanation:

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Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

or

b > 3\frac{2}{15}\\\\

========================================================

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality 2 \frac{3}{5} < b - \frac{8}{15}\\\\ is the same as \frac{13}{5} < b - \frac{8}{15}\\\\

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

Let's multiply both sides by 15 to clear out the fractions

\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\

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

Now isolate the variable b

39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

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