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marin [14]
3 years ago
9

If 5 over 6 gallon covers 1 over 4 of the house then how much paint is needed for the entire house

Mathematics
1 answer:
MissTica3 years ago
4 0
If 5/6 gallons covers a quarter of the house, then you would need 4 times as much paint:
\frac{5}{6}  \times 4
Which is 20/6 gallons or 3 and 1/3 gallons
Hope this helped
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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
∠3 and ∠6 can be classified as:
anastassius [24]

Answer:

Alternate Interior Angles

Step-by-step explanation:

Since they are inside the parallel lines, Alternate Exterior Angles and any other similar theorems can be ruled out.

Since they are on opposite sides of each other, Corresponding Angles and any other similar theorems can be ruled out.

Since they are far apart from each other, Supplementary Angles, Adjacent Angles, Vertical Angles, and any other similar definitions can be ruled out.

Therefore, we are left with Alternate Interior Angles.

6 0
3 years ago
Read 2 more answers
If triangle RST = triangle NPQ, then RT is congruent to ?. . a.NP. b.NQ. c.PQ. d.QP
lapo4ka [179]
If the triangles are given to be congruent, the corresponding angles and corresponding sides should have the same measure and lengths. From the given, side RT is congruent to side NQ. Therefore, the answer to the question is letter B. NQ. 
6 0
2 years ago
Mary’s coffee shop borrowed $9,850 for new chairs and tables. The money was borrowed for 9 months at 9 ¼ % discounted rate, what
Greeley [361]
Let
M---------------> money borrowed -------------> <span>$9,850
r--------------> </span>discounted rate--------> <span>9 ¼=9.25-------> 0.0925
t---------------> time--------> </span><span>9 months=9*30=270 days
D-------------> </span><span>amount of the discount

we know that
D=M*r*t/360=(9850)*(0.0925)*(270/360)=683.34

the answer is $683.34</span>
6 0
3 years ago
A chemist mixes a 10% hydrogen peroxide solution with a 25% hydrogen peroxide solution to create 30 liters of a 15% hydrogen per
kupik [55]
The answer is 20 liters.
6 0
3 years ago
Read 2 more answers
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