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4 years ago

WHATS 4 + 2 + 10 + 5

Mathematics

2 answers:

olganol [36]4 years ago

6 0

Answer:

Step-by-step explanation:

haha funny number

Luba_88 [7]4 years ago

3 0

The answer to this problem is 21

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The variable q represents a positive integer. Does 3(59 + 9) + 5q represent a number that is greater than, less than, or equal t

hjlf

Answer:

Depends on the value of q.

Step-by-step explanation:

3(59+9) + 5q = 3(68) + 5q = 204 + 5q

4(59+6) = 260

7 0

3 years ago

I’m really confused

shtirl [24]

Answer:

You've got it correct, if I'm not mistaken it's either B, or C. I don't remember which answer it was, it was one of those 50/50.

5 0

3 years ago

How do you set up and solve these inequalities-only 16-20

xxMikexx [17]

Sorry I'm late, here are the equations, just solve for x! Treat the inequalities like an = sign and solve!!

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

3 years ago

What is the approximate volume of one refraction cup? What’s its formula ?

Marizza181 [45]

Answer:

Volume = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup

Step-by-step explanation:

The dimensions of a refraction cup are;

Diameter = 120 mm

Depth = 30 mm

Therefore, volume of the refraction cup = Area of base × Depth

Volume of the refraction cup = (π×120²/4)/2 × 30 = 169646.003 mm³

1 L = 1000000 mm³

Therefore, 169646.003 mm³ = 169646.003/1000000 m³ = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup.

6 0

3 years ago