I don’t know how to explain on why the student was incorrect

Mathematics

1 answer:

Zolol [24]2 years ago

7 0

Answer:

the constant proportionality means the ratio between two directly proportional quantities.

so the constant proportionality will be 7.5÷6=1.25 or 10÷8=1.25 or 12.5÷10=1.25

constant proportionality should be same all through because it's a constant

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yolanda had 47 customers on saturday from 6pm to 7pm on monday from 5 pm to 6 pm she had 20 customers what is the amount of perc

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3 years ago

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Serjik [45]

The Pythagorean's Theorem for our situation would look like this:

$shortleg^2+longleg^2=hypotenuse^2$

So let's call the short leg s, the long leg l and the hypotenuse h. It appears that all our measurements are based on the measurement of the short leg. The long leg is 4 more than twice the short leg, so that expression is l=2s+4; the hypotenuse measure is 6 more than twice the short leg, so that expression is h=2s+6. And the short leg is just s. Now we can rewrite our formula accordingly:

$s^2+(2s+4)^2=(2s+6)^2$

And of course we have to expand. Doing that will leave us with

$s^2+4s^2+8s+8s+16=4s^2+12s+12s+36$

Combining like terms we have

$5s^2+16s+16=4s^2+24s+36$

Our job now is to get everything on one side of the equals sign and solve for s

$s^2-8s-20=0$

That is now a second degree polynomial, a quadratic to be exact, and it can be factored several different ways. The easiest is to figure what 2 numbers add to be -8 and multiply to be -20. Those numbers would be 10 and -2. Since we are figuring out the length of the sides, AND we know that the two things in math that will never EVER be negative are time and distance/length, -2 is not an option. That means that the short side, s, measures 10. The longer side, 2s+4, measures 2(10)+4 which is 24, and the hypotenuse, 2s+6, measures 2(10)+6 which is 26. So there you go!

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3 years ago

Explain whether the following given sets are closed under addition:

Ivan

Answer:

Step-by-step explanation:

a) Yes.

$\forall\ a,b\ \in\ B,\ a+b\in B\ (since\ 0+0=0)$

b) Yes

$W=\{0,1,2,3,..,\}\ (whole\ numbers)\\\forall\ a,b\ \in\ 3W,\ a+b\in 3W\\since:\\a\ \in\ 3W \Longrightarrow\ \exists\ k_1\in W | a=3*k_1\\\\b\ \in\ 3W \Longrightarrow\ \exists\ k_2\in W | b=3*k_2\\\\a+b=3*k_1+3*k_2=3*(k_1+k_2) \in 3W\\$