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Hitman42 [59]
3 years ago
6

3a. Write an equation in slope-intercept form of a line that passes through (2,1) and (6,-5).

Mathematics
1 answer:
eimsori [14]3 years ago
5 0

Answer:

y =- 3/2x + 4

Step-by-step explanation:

(2,1) and (6,-5).\\x_1 = 2\\x_2 = 6\\y_1 =1\\y_2 =-5\\\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\\ \\\frac{y-1}{x-2} = \frac{-5-1}{6-2}\\\\\frac{y-1}{x-2} = \frac{-6}{4} \\Cross-Multiply\\4(y-1) = -6(x-2)\\4y-4=-6x+12\\4y =-6x+12+4\\4y = -6x+16\\Divide-  through-by ; 4\\\frac{4y = -6x+16}{4} \\\\y = -\frac{3}{2} x +4

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Help please with this question
N76 [4]

Answer:

The third graph down

Step-by-step explanation:

-2.1 w > 12.81

Divide each side by -2.1, remembering to flip the inequality

-2.1w / -2.1 < 12.81/ -2.1

w < -6.1

There is an open circle at -6.1 and the line goes to the left

6 0
3 years ago
Read 2 more answers
For the function defined by f(t)=2-t, 0≤t&lt;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
It takes 32 hours for a motorboat moving downriver to get from pier A to pier B. The return journey takes 48 hours. How long doe
lina2011 [118]

Answer:

Time taken by the un powered raft to cover this distance is T = 192.12 hr

Step-by-step explanation:

Let speed of boat = u \frac{km}{hr}

Speed of current = v \frac{km}{hr}

Let distance between A & B = 100 km

Time taken in downstream = 32 hours

32 = \frac{100}{u +v}

u + v = \frac{100}{32}

u + v = 3.125 ------ (1)

Time taken in upstream = 48 hours

48 = \frac{100}{u -v}

u -v = \frac{100}{48}

u - v = 2.084 ------- (2)

By solving equation (1) & (2)

u = 2.6045 \frac{km}{hr}

v = 0.5205 \frac{km}{hr}

Now the time taken by the un powered raft to cover this distance

T = \frac{100}{v}

Because un powered raft travel with the speed of the current.

T = \frac{100}{0.5205}

T = 192.12 hr

Therefore the  time taken by the un powered raft to cover this distance is

T = 192.12 hr

3 0
3 years ago
Read 2 more answers
2 d2 + 15d + 23 = 3<br> It’s quadratic formula
DENIUS [597]

Answer:

-15 plus or minus the square root of 65 all over 4 (All over means fraction)

4 0
3 years ago
110% of 90 is what number a ?<br><br>? = 110 over 100<br>please help​
densk [106]

Answer:

90 is the orrect answer because I said so

Step-by-step explanation:

3 0
2 years ago
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