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2 years ago

The grade six pupils decided to have an income generating project. They make

Mathematics

1 answer:

lara31 [8.8K]2 years ago

8 0

Answer:

c. P 2.40

Step-by-step explanation:

P 100 spent, 50 pieces made. Therefore P2 a piece. 20% of P2 is .40. Therefore total is 2 + .40 = P2.40

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Combine like terms and note the degree 8a^6bc-abc-8a^6bc

erik [133]

Answer:-abc

Step-by-step explanation:

To collect like terms of

8a^6bc-abc-8a^6bc

=8a^6bc-8a^6bc-abc

=-abc

7 0

3 years ago

A rectangular box is 6 inches long, 5 inches wide and 7 inches tall. What is the approximate length of its longest diagonal?

xxTIMURxx [149]

Answer:

10.5 inches

Step-by-step explanation:

The length of the longest diagonal is across two opposite points of the solid, namely

L = sqrt ( 6^2 + 5^2 + 7^2)

= sqrt (36+25+49)

= sqrt(110)

= 10.5"

7 0

3 years ago

Read 2 more answers

2.6432 rounded to the nearest tenth

AlexFokin [52]

Answer:

2.6

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Which is the angle of depression from b to c

matrenka [14]

Step-by-step explanation:

that the answer. because it below par

8 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