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klio [65]
3 years ago
7

Help please.............

Mathematics
2 answers:
Aleksandr [31]3 years ago
5 0

Answer:

10 C

Step-by-step explanation:

Kruka [31]3 years ago
4 0

Answer: D Or 10

Step-by-step explanation: Because If you subtract 50 by 32 you will get 18 and then you get 5/9(18) and then you can cross simplify which gets you 5x2 which is 10.

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Find the exact value of sin(u+v) given that sin u=7/25 and cos v=-12/13
Helen [10]

Answer

Step-by-step explanation:

4 0
3 years ago
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The Kitchen committee purchased 76 boxes of cookies for Vacation Bible School,
iren2701 [21]
269 I think :) not sure
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2 years ago
Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?
Sergeu [11.5K]
Hello!

First of all we find the slope by dividing the difference of the y values by the difference of the x values as seen below.

\frac{-3-11}{1+3} = -\frac{14}{4} = \frac{7}{2}

The slope of our line is 7/2, or 3.5
-------------------------------------------------------------
Now we will put the slope and a point on our line into slope intercept form and solve for b. We will use (-3,11).

11=3.5(-3)+b
11=10.5+b
b=0.5

Our final equation is shown below.

y=3.5x+0.5

I hope this helps!


7 0
3 years ago
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What is the area of the two-dimensional cross section that is parallel to face ABC ?
Afina-wow [57]
The two triangle faces are congruent, so the missing side length on the top triangle is 24ft.
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8 0
3 years ago
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Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e
Bingel [31]
Assuming the series is

\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}

The series will converge if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|

We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|

So the series will certainly converge if -1, but we also need to check the endpoints of the interval.

If x=1, then the series is a scaled harmonic series, which we know diverges.

On the other hand, if x=-1, by the alternating series test we can show that the series converges, since

\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0

and is strictly decreasing.

So, the interval of convergence for the series is -1\le x.
6 0
3 years ago
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