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Paraphin [41]
3 years ago
5

How do i do x-y coordinates

Mathematics
2 answers:
mestny [16]3 years ago
7 0
You have to write it as (x,y)
const2013 [10]3 years ago
6 0
You have to put them down as x-y
You might be interested in
Simplify 3/4(1/2*-12)+4/5
Lyrx [107]
The exact form is:
-37/10


The decimal for is:
-3.7


The mixed number for is:
-3 7/10
7 0
3 years ago
The average number of points a basketball team scored for
Ganezh [65]
Average is

(2x + X+6)/3=63
(2x+x+6)= 189
3x+6=189
3x=183 61

X=61
Scores for the game are: 61, 61, and 67
3 0
3 years ago
Multiply and simplify if possible
ad-work [718]

Answer:

6 √ ( 3 x + 5 ) ( 3 x − 2 )

Step-by-step explanation:

3 0
3 years ago
Carmen is painting a wall that is 26 feet by 16 feet. What is the perimeter on the wall , in inches?
strojnjashka [21]
26×2=52
16×2=32
52+32=84
Perimeter=84
8 0
2 years ago
Read 2 more answers
Determine whether the integral is convergent or divergent. [infinity] 7 e−1/x x2 dx convergent divergent Changed: Your submitted
inessss [21]

I suppose the integral could be

\displaystyle\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx

In that case, since -\frac1x\to0 as x\to\infty, we know e^{-1/x}\to1. We also have \left(e^{-1/x}\right)'=\frac{e^{-1/x}}{x^2}>0, so the integral is approach +1 from below. This tells us that, by comparison,

\displaystyle\frac{e^{-1/x}}{x^2}\le\frac1{x^2}\implies\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx\le\int_7^\infty\frac{\mathrm dx}{x^2}

and the latter integral is convergent, so this integral must converge.

To find its value, let u=-\frac1x, so that \mathrm du=\frac{\mathrm dx}{x^2}. Then the integral is equal to

\displaystyle\int_{-1/7}^0e^u\,\mathrm du=e^0-e^{-1/7}=1-\frac1{\sqrt[7]{e}}

4 0
3 years ago
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