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3 years ago

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A 220-watt microwave is used an average of two hours a day. Find the cost of using the microwave for two weeks at a cost of $0.1

2 per kilowatt-hour. $105.60 $0.74 $0.11 $739.20

2 answers:

HACTEHA [7]3 years ago

6 0

Answer: Its $739.20

Step-by-step explanation:

Take 220 and multiply it by the 2 hours. Then take the product of that and multiply it by 14-2 weeks. After that, do the product of that times 0.12 and you have your answer. Hope that helped. Let me know of you got confused.

Alex73 [517]3 years ago

4 0

Answer: $0.74

Step-by-step explanation: First, find the watts used per day. 220 x 2 = 440 Wh. Then find the watt-hours used for two weeks. 440 Wh x 14 = 6160 Wh. Convert Wh to kWh. 6160 Wh (1 kWh/1000 Wh) = 6.16 kWh. Calculate the cost. 6.16 kWh x $0.12 = $0.74

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i will give brainliest.The product of 5 and the sum of 12 and a certain number is 10. What is the number

rjkz [21]

Answer: x=-10

Step-by-step explanation:

5*(12+x)=10

Distribute

60+5x=10

Subtract(60)

5x=-50

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<em>Hope it helps <3</em>

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3 years ago

Read 2 more answers

Use trigonometric ratios to find the indicated side

iragen [17]

Answer:

a = 139.1

b = 56.2

Step-by-step explanation:

A. Reference angle = 68°

Opp = a

Hyp = 150

Therefore:

Sin 68 = opp/hyp

Sin 68 = a/150

150*sin 68 = a

a = 139.1 (nearest tenth)

B. Reference angle = 68°

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2 years ago

The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

$\frac{10000(0.2)^1}{10000(0.2)^0}$

$\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2$

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

8 0

3 years ago

An observer (O) is located 500 feet from a school (S). The observer notices a bird (B) flying at a 39° angle of elevation from h

Serhud [2]

Answer:

h=404.89

Step-by-step explanation:

We are looking for side h, which is opposite of the observer. We know that the side adjacent to the observer is 500 feet. We also know that the angle from the observer to the bird is 39°. Because we have these values, SOH CAH TOA tells us that we should use tangent, opposite over adjacent.

We can set up our equation as follows:

tan(39°)=h/500

We can then solve for h:

500*tan(39°)=h

h=404.89

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2 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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3 years ago