All categories

3 years ago

Expand the following bracket -5(3c+6)

Mathematics

2 answers:

wel3 years ago

6 0

Answer:

The answer is -15c - 30

Step-by-step explanation:

You have to apply Distributive Law :

$a(m + n) = am + an$

So for this question :

$- 5(3c + 6)$

$= - 5(3c) - 5(6)$

$= - 15c - 30$

Lynna [10]3 years ago

5 0

Answer:

-15c - 30

Step-by-step explanation:

-5(3c+6)

Expand or distribute the term outside the bracket to the terms inside.

-5(3c) - 5(6)

-15c - 30

You might be interested in

What is X equal to? It’s not 180 -.-

Natali5045456 [20]

Answer:

x = 163

Step-by-step explanation:

(x-5)+22= 180

x = 143

hope this helps!!!

3 0

3 years ago

Read 2 more answers

Which quadratic equation has roots - 2 and 1/5 :

Sladkaya [172]

Answer:

$5 {x}^{2} \: + \: 8x \: - 2 \: = \: 0$

Step-by-step explanation:

$sum \: of roots = - 2 + \frac{1}{5} = - \frac{8}{5} \\ product \: of \: roots = - 2 \times \frac{1}{5} = - \frac{2}{5} \\ from \: general \: equation \\ a {x}^{2} + bx + c = 0 \\ {x}^{2} + \frac{8}{5} x \: - \frac{2}{5} = 0 \\ 5 {x}^{2} \: + \: 8x \: - 2 \: = \: 0$

7 0

3 years ago

The region in the first quadrant bounded by the x-axis, the line x = ln(π), and the curve y = sin(e^x) is rotated about the x-ax

charle [14.2K]

First, it would be good to know that the area bounded by the curve and the x-axis is convergent to begin with.

$\displaystyle\int_{-\infty}^{\ln\pi}\sin(e^x)\,\mathrm dx$

Let $u=e^x$ , so that $\mathrm dx=\dfrac{\mathrm du}u$ , and the integral is equivalent to

$\displaystyle\int_{u=0}^{u=\pi}\frac{\sin u}u\,\mathrm du$

The integrand is continuous everywhere except $u=0$ , but that's okay because we have $\lim\limits_{u\to0^+}\frac{\sin u}u=1$ . This means the integral is convergent - great! (Moreover, there's a special function designed to handle this sort of integral, aptly named the "sine integral function".)

Now, to compute the volume. Via the disk method, we have a volume given by the integral

$\displaystyle\pi\int_{-\infty}^{\ln\pi}\sin^2(e^x)\,\mathrm dx$

By the same substitution as before, we can write this as

$\displaystyle\pi\int_0^\pi\frac{\sin^2u}u\,\mathrm du$

The half-angle identity for sine allows us to rewrite as

$\displaystyle\pi\int_0^\pi\frac{1-\cos2u}{2u}\,\mathrm du$

and replacing $v=2u$ , $\dfrac{\mathrm dv}2=\mathrm du$ , we have

$\displaystyle\frac\pi2\int_0^{2\pi}\frac{1-\cos v}v\,\mathrm dv$

Like the previous, this require a special function in order to express it in a closed form. You would find that its value is

$\dfrac\pi2(\gamma-\mbox{Ci}(2\pi)+\ln(2\pi))$

where $\gamma$ is the Euler-Mascheroni constant and $\mbox{Ci}$ denotes the cosine integral function.