I need to know how to solve this and what is the question asking

Mathematics

1 answer:

expeople1 [14]3 years ago

8 0

You can reword the equation 6x+y=9 to
Y=6x+9

Then plug-in y=6x+9 into the equation 2x+3y=-5

Plug this 2x+3(6x+9)=-5
2x+18x+27=-5
Then solve it from there

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Helppppp how do u do this????

otez555 [7]

Lets x to be number of students.

1) If each students get 3 ml, then all students get 3x ml.
There was n ml.
(n-3x) ml leftover

n - 3x = 5

2) If each students get 4 ml.
then all students get 4x ml.

n+21 = 4x

3)
n - 3x = 5
- (n+21 = 4x) -------> -n -21=-4x

n - 3x = 5
-n -21= - 4x
-3x-21=5 - 4x
x=5+21
x=26

Answer : 26 students

Check:
x=26, n-3x = 5, n-3*26=5, n=83
n+21 = 4x
83+21=4*26
104 = 104 true

7 0

3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

2 years ago

Mr. Scott’s workroom is a rectangle that measures 25 feet by 27 feet. His woodworking equipment takes up an area of 300 ft squar

Mashutka [201]

375 because if you do 27x25=675 then I did 675-300=375.
So if my math is correct 375 feet is not covered by the woodworking equipment

7 0

3 years ago

Read 2 more answers

How would I solve this?

soldi70 [24.7K]

6-2x=6x-10x+20 solve for x

combine like terms and get all to the same side of equation
6-2x+10x-6x=20
10x-2x-6x+6=20
8x=20-6
8x=14
x=14/8=7/4

6 0