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2 years ago

The sum of twice a number and five is greater than three times the number minus three.

Mathematics

2 answers:

AnnZ [28]2 years ago

5 0

Answer:

x<8

Step-by-step explanation:

2x+5>3x-3

2x+8>3x

8>x

strojnjashka [21]2 years ago

5 0

Answer:

n < 8

Step-by-step explanation:

Let n represent the unknown number. Then:

2n + 5 > 3n - 3

Collecting like terms:

8 > n or n < 8

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2 years ago

Read 2 more answers

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.

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2 years ago

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3 years ago

How many unique values of x solve x2 + 14x + 49 = 0?

Natalka [10]

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2 years ago

1. Simplify. 5–1(3–2)

konstantin123 [22]

Answer with explanation:

Using following law of indices

$1.x^a \times x^b=x^{a+b}\\\\2. (x^m)^n=x^{mn}\\\\ 3. x^{-n}=\frac{1}{x^n}\\\\4. \frac{x^m}{x^n}=x^{m-n}$

$1.5^{-1}\times 3^{-2}=\frac{1}{5}\times\frac{1}{3^2}\\\\=\frac{1}{5\times9}\\\\=\frac{1}{45}\\\\2. \frac{mn^{-4}}{p^0*q^{-2}}=\frac{mq^2}{n^4}\\\\ 3.0.0042=4.2\times 10^3\\\\4.6.12 \times 10^3=6.12 \times 1000=6120\\\\ 5.0.5 \times(8\times10^5})=\frac{1}{2}\times8 \times 10^5=4\times 10^5\\\\6. (9 \times 10^3)^2=9^2 \times (10^3)^2=81 \times 10^6\\\\7.( \frac{1}{2}a^{-4}\times b^2 )_{a=-2,b=4}{\text{or}}( \frac{1}{2}a^{4}\times b^{-2} )_{a=-2,b=4}=\frac{1}{2}\times (-2)^{-4}\times(4)^2{\text{or}}=\frac{1}{2}\times (-2)^{4}\times(4)^{-2}=\frac{16}{2\times16}=\frac{1}{2}$

$9. (4\times x\times y^2)^3\times (xy)^5=4^3\times x^3\times (y^2)^3\times x^5 \times y^5\\\\=64\times x^{3+5}\times y^{5+6}\\\\=64x^8y^{11}\\\\10.(2\times t^3)^3(0.4\times r)^2=2^3 \times (t^3)^3\times (0.4)^2 \times r^2=8\times t^9 \times 0.16 \times r^2=1.28r^2t^9$