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VikaD [51]
3 years ago
7

Help!!!!!!!! Plz I need help

Mathematics
1 answer:
GarryVolchara [31]3 years ago
7 0

Answer:

2 and 3

Step-by-step explanation:

Hope this helps.

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Sandy works at a clothing store.She makes $8 per hour plus 10% commission on her sales.She worked 79 hours over the last two wee
Nonamiya [84]
873 for sure 2,414x10%=241 + 8x79=632 =873
5 0
3 years ago
Roll a dice. Chance of it landing on a 6 is ? In ?
ohaa [14]
It is a 1 to 6 ratio
4 0
4 years ago
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What is the soloution to -4x+2=-14​
docker41 [41]

Answer:

x = 4

Step-by-step explanation:

subtract 2 from both sides. then -4x is equal to -16. divide by -4 which gives you 4. Hope this helped!

3 0
3 years ago
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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).
4 0
3 years ago
What is the surface area of a sphere with a circumference of 100T? Leave
mixer [17]

Answer:

10000π units²

Step-by-step explanation:

From the question given above, the following data were obtained:

Circumference (C) = 100π

Surface Area of sphere (SA) =?

Next, we shall determine the radius. This can be obtained as follow:

Circumference (C) = 100π

Radius (r) =?

C = 2πr

100π = 2πr

Divide both side by 2π

r = 100π / 2π

r = 50 units

Finally, we shall determine the surface area of the sphere. This can be obtained as follow:

Radius (r) = 50 units

Surface Area of sphere (SA) =?

SA = 4πr²

SA = 4 × π × 50²

SA = 4 × π × 2500

SA = 10000π units²

Therefore, the surface area of the sphere is 10000π units²

3 0
3 years ago
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