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olya-2409 [2.1K]
3 years ago
9

An onion soup recipe calls for 3 2/3 cups of chopped onions. Katrina has already chopped 1 1/3 cups of onions. She wants to know

how many more cups she needs to chop. Katrina still needs to chop cups of onion soup recipe calls for 3 2/3 cups of chopped onions. Katrina has already chopped 1 1/3 cups of onions. She wants to know how many more cups she needs to chop. Katrina still needs to chop cups of onions
Mathematics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

2\dfrac{1}{3}\ \text{cups}

Step-by-step explanation:

Amount of chopped onions required = 3\dfrac{2}{3}\ \text{cups}

Amount of onions Katrina has already chopped = 1\dfrac{1}{3}\ \text{cups}

Amount of onions that are required to be chopped is given by

3\dfrac{2}{3}-1\dfrac{1}{3}\\ =\dfrac{11}{3}-\dfrac{4}{3}\\ =\dfrac{7}{3}\\ =2\dfrac{1}{3}\ \text{cups}

Hence, the amount of onions that are required to be chopped is 2\dfrac{1}{3}\ \text{cups}.

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To find the inverse of the function, change f(x)=x+3
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Answer:

Step-by-step explanation:

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3 years ago
a rental car company charges drivers a flat fee to rent a car plus $0.25 for each mile they drive. If a drivers pays $100 drives
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ANSWER

y = 100 + 0.25(160) - equation

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2 years ago
When a teacher divided her students into groups of four, she had three students remaining. when she divided them into groups of
AfilCa [17]
39 students....39/4 = 9 remainder 3
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3 years ago
How to know if a function is periodic without graphing it ?
zhenek [66]
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).
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Remember, the slope formula is y2-y1 over x2-x1. To plug numbers into the formula, find two points. Then, you substitute the points into the formula.
3 0
3 years ago
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