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In-s [12.5K]
4 years ago
10

What is seven times the quotient five and seven

Mathematics
1 answer:
sdas [7]4 years ago
6 0


7 x [5/7) = 5

The sevens cancel out each other leaving an answer of 5.

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A, B, and C are collinear points. B is between A and C. AB = 5x + 8 BC = 6x - 1 AC = 12x - 11 Find AC.
Vlada [557]

Answer:

AC = 198

Step-by-step explanation:

Since all these points are collinear, we know that the addition of AB plus BC should give the same as AC. We can then set an equation that addresses this identity:

AB + BC = AC

5x +8 +6x - 1 = 12x - 11

Now re-arranging like terms in order to combine them:

8 - 1 + 11 = 12x - 5x - 6x

19 - 1  = 12x - 11x

18 = x

Now that we know the value of 'x", we can determine the value of AC:

AC = 12x - 11

AC = 12 (18) - 11

AC = 216 - 18

AC = 198

8 0
4 years ago
Rachel has 37 videos and decides to purchase 2 more each week. Write an equation describing this situation.
sergiy2304 [10]
Answer:

T = 2w + 37

Step-by-step explanation:

Let T represent the total amount of videos Rachel has after w weeks

Rachel already owns 37 videos

She purchases 2 more each week

Therefore, after w weeks, the total amount of videos Rachel has can be represented by the expression:

2w + 37

Then the total after w weeks can be modeled by the equation:

T = 2w + 37

8 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
4 years ago
The length of a rectangle is seven less than twice the length of its width. If the area of the rectangle is 15 square meters, fi
Brums [2.3K]

Answer:

Length is 3 meters and the Width is 5 meters

Step-by-step explanation:

L*W=15

L=2W-7

(2W-7)W=15

2W^{2} -7W=15\\2W^{2} -7W-15=0

(2W+3)(W-5)=0, don't use (2W+3)=0, because it gives you a negative dimension.

W=5, L=3

4 0
3 years ago
The area of a parallelogram is 216 cm². The height of the parallelogram is 2/3 of its base. Find the base and height.
grin007 [14]
2 equations, 2 unknowns. Call height H and base B.

From the first sentence:
HB = 216

From the second sentence:
H = ⅔ B

Plug the second into the first and solve for B:
(⅔ B)B = 216
B^2 = 324
B = 18

Plug that back into either of the original 2 equations and solve for H:
H(18) = 216
H = 12
3 0
3 years ago
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