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Alexandra [31]
2 years ago
14

Apply the distributive property to simplify the expression-4(5x +2)

Mathematics
2 answers:
Art [367]2 years ago
8 0

Answer:

-20x-8

Step-by-step explanation:

Apply the distributive property: a(b + c) = ab + ac

Sedaia [141]2 years ago
7 0

Answer:

−20x − 8

Step-by-step explanation:

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Sarah opens a small bag of chocolate candies and notices that out of 210 candies, 63 are brown. She assumes the color breakdown
mezya [45]

Answer:

1440

Step-by-step explanation:

We can solve this problem by applying the rule of three.

In fact, we know that:

- Over a package of 210 candies,

- The number of brown candies is 63

- Here we want to find what is the number of brown candies when the total number of candies contained in the package is 4800

So we can set up the following rule of three:

\frac{x}{4800}=\frac{63}{210}

where

x = number of brown candies when the total number of candies contained in the package is 4800

Solving the expression for x, we find:

x=\frac{63}{210}\cdot 4800 =1440

So, Sarah can expect to find 1440 brown candies in a package of 4800 pieces.

7 0
3 years ago
For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
3 years ago
Nasrin invested $6000 in a bond at a yearly rate of 3%. She earned $450 in interest. How long was the money invested?​
Alenkasestr [34]
4 years

6000 x .3 = 1,800. 1800 divided by 450 = 4
6 0
2 years ago
Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a drought occurs, the prob
astraxan [27]

Answer:

the correct option is d. 0.02

Step-by-step explanation:

We have the following events:

x: A drought occurd

y: Water rationing.

Therefore, according to the plaster we have:

P [x] = 0.20

P [y | x] = 0.10

Now P [it is a drought and water rationing happens] =

P [x n y] = P [y | x] * P [x] = 0.10 * 0.20 = 0.02

Which means that the correct option is d. 0.02

3 0
3 years ago
Please answer this as soon as possible!! Please and thank you!
salantis [7]

Step-by-step explanation:

Find w

V = lwh

V ÷ lh = w

V/lh = w

w = V/lh

Option → C

3 0
2 years ago
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