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GREYUIT [131]
3 years ago
5

Y=1/2x+4 I need to help solving this!! :)

Mathematics
2 answers:
sweet [91]3 years ago
7 0

Answer: so the 4 is the y intercept and thats where u plot one point (0,4) then 1/2x is the slope so its rise/run so it goes up 1 and goes over 2 squares. so like the second point you would plot is (2,5) then you draw the line

grandymaker [24]3 years ago
7 0

Answer:

put the point on 4 going up but dont go over so just put one point on the 4 on the y axis. And then put another point over 2 and then up 5 but make sure you start in the middle. ask me if you need help

Step-by-step explanation:

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How do I work this out??please help
serg [7]
The answer is x=5 because 24-3-5=16 so you have to get those 2 sides to equal 16....2*5=10+1=11 then add 5 and you get 16
4 0
4 years ago
Read 2 more answers
Don’t know the answer need help
Romashka [77]
Answer: The point was reflected over the x-axis

Explanation:
When the y coordinate is opposite, then it’s mirror on the x-axis. If the x was opposite, then it would have been over the y-axis.
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3 years ago
Find the equation of the line in slope-intercept form.Slope is 2/3 through (3, 4)
Lena [83]

Answer:

y = (2/3)x + 2

Step-by-step explanation:

Start with the general slope-intercept form y = mx + b.  Substitute 3 for x and 4 for y and 2/3 for m, and find the intercept, b:

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3 0
4 years ago
Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1
erma4kov [3.2K]

a. The series is

\displaystyle\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-4-\frac43-\frac4{3^2}-\frac4{3^3}-\cdots

(first four terms are listed)

b. The series converges because this is a geometric series with r=\dfrac13.

c. Let S_N be the N-th partial sum of the series:

S_N=\displaystyle\sum_{n=1}^N-4\left(\frac13\right)^{n-1}

S_N=-4-\dfrac43-\dfrac4{3^2}-\cdots-\dfrac4{3^{N-1}}

Multiplying both sides by \dfrac13 gives

\dfrac13S_N=-\dfrac43-\dfrac4{3^2}-\dfrac4{3^3}-\cdots-\dfrac4{3^N}

Subtracting this from S_N gives

S_N-\dfrac13S_N=\dfrac23S_N=-4+\dfrac4{3^N}

\implies S_N=-6+\dfrac6{3^N}

As N gets larger and larger (N\to\infty) the rational term converges to 0 and we're left with

\displaystyle\lim_{N\to\infty}S_N=\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-6

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3 years ago
PLEASE SOMEONE ASAP HELP ME!!!!!!!!!!!
MrRissso [65]
She distributed incorrectly.
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3 years ago
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