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2 years ago

What should be the first step to solve the equation?

Mathematics

2 answers:

katrin [286]2 years ago

7 0

Answer:

Step-by-step explanation:2x + 9x - 9 = 16x + 16 - 5

= 11x - 9 = 16x + 11

11x - 16x = 11 + 9

5x = 20

x = 20/5

x = 4

Plz mark it as brainliest:)

4vir4ik [10]2 years ago

7 0

Answer:

Step-by-step explanation:

Use the distributive property on each side of the equation.

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Please help marking brainliest if it's correct and explain

yarga [219]

Imma take a guess maybe C

4 0

3 years ago

Can you guys give me a serious answer please?

grin007 [14]

Answer:

800 bacteria

Step-by-step explanation:

Answer of this question depend upon number of hours at the end of day

Let Take Full Day of 24 hrs

Colby bacteria will doubled 12 times ( doubled once in 2hrs => 24/2 = 12)

Colby bacteria at end of day = 50 * 2¹²

Jaquan bacteria will doubled 8 times ( doubled once in 3hrs => 24/3 = 8)

Let say Jaquan bacteria at start of Day = J

Jaquan bacteria at end of day = J * 2⁸

Equating Both

J * 2⁸ = 50 * 2¹²

=> J = 50 * 2⁴

=> J = 50 * 16

=> J = 800

Jaquan will start with 800 Bacteria

4 0

1 year ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2

Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

7 0

1 year ago

Please help 15 points each but beware if you use my points for no reason I'll report you

romanna [79]

Answer:

$2048~ \text{in}^3$

Step-by-step explanation:

$\text{Volume,}~ V = \dfrac 43 \pi r^3 \\\\\\~~~~~~~~~~~~~~~=\dfrac 43 \times 3 \times 8^3\\\\\\~~~~~~~~~~~~~~~=4 \times 512\\\\\\~~~~~~~~~~~~~~~=2048~ \text{in}^3$