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nasty-shy [4]
3 years ago
10

Aiden has a goal to lose 5% of his current body weight. If Aiden's current weight is w pounds, which expression does NOT represe

nt Aiden's goal weight?
A. w - 0.05w
B. 1.00w - 0.05w
C. (1.00 - 0.05)w
D. 1.00 - 0.05w
Mathematics
1 answer:
butalik [34]3 years ago
3 0
Hello! So, with Adian having a goal to lose 5% of his body weight, that means he will still have 95% of it left. The variable "w" represents the current amount of his weight. Multiplying by 1 will still give you his weight. Suppose he weighs 200 pounds. You plug in that as "w". Then, you would multiply that number by 5% (0.05) to see how much he would need to lose. 200 * 0.05 is 10. 200 - 10 is 190. Multiplying any number by 1 will get you to the same number. B and C are out, because they are basically the same equation, even when the "w" in answer choice C is distributed. A is out, because it shows the way the problem would be set up. D does not represent, because 1 is not his weight and it does not give a realistic answer on how much he needs to lose and weigh. Therefore, the answer is D.
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Which situation could best be represented by the integer 120
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2 years ago
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lisov135 [29]

Answer:

a

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3 years ago
How to know if a function is periodic without graphing it ?
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A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

The second two are satisfied whenever k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}, and more generally when k=\dfrac{10n}7 with n\in\mathbb Z (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5

More generally, it can be shown that

f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))

is periodic with period \mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n).
4 0
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A 17.2 gallon gasoline tank is 1/2 full. How many gallons will it take to fill the tank?
love history [14]

Answer:

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Step-by-step explanation:

We know that the tank is filled halfway, and it can hold 17.2 gallons, so we can just divide to figure out how much is half the tank (the amount we need to fill it)

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We can check our work by multiplying

8.6 x 2 = 17.2

So it would take 8.6gal to fill it up. I hope this helps :)

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