All categories

2 years ago

Simplify 1/3n (-6+27m-51p)

Mathematics

2 answers:

marishachu [46]2 years ago

7 0

1/3n * -6 = -2n
1/3n * 27m = 9nm
1/3n* -51p = -17np
-2n + 9nm - 17np

Alinara [238K]2 years ago

7 0

$\frac{1}{3}n(-6+27m-51p)=-6* \frac{1}{3}n +27m* \frac{1}{3}n-51p* \frac{1}{3}n= \\ \\ =-2m+9mn-17np$

You might be interested in

Determine which of the lines, if any, are parallel or perpendicular. Explain.

jonny [76]

Line A. line B. line C. it’s A

8 0

3 years ago

A drum has a diameter of 10 inches. find the area of the top of the drum. use 3.14 for pi.

icang [17]

Answer:

My answer is 78.55

Step-by-step explanation:

I've given the steps. Hope it really helps

4 0

2 years ago

Read 2 more answers

The amount of money invested or borrowed is known as:

devlian [24]

B:Interest

Hoped this helped have a nice day

8 0

2 years ago

Read 2 more answers

Divide.<br> a.1 − 2i/2i<br> b. 5 − 2i/5 + 2i<br> c.√3 − 2i/−2 − √3i

Umnica [9.8K]

Answer:

(a) \frac{-1i}{2}-1[/tex]

(b) $\frac{20i+21}{29}$

Step-by-step explanation:

We have to perform division

(a) $\frac{1-2i}{2i}$

So after division

$\frac{1-2i}{2i}=\frac{1}{2i}-\frac{2i}{2i}=\frac{-1i}{2}-1$

(b) We have given expression $\frac{5-2i}{5+2i}$

After rationalizing $\frac{5-2i}{5+2i}\times \frac{5-2i}{5-2i}=\frac{(5-2i)^2}{25+4}=\frac{25+4i^2+20i}{29}=\frac{20i+21}{29}$

After rationalizing

$\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}\times \frac{-2+\sqrt{3i}}{-2+\sqrt{3i}}=\frac{-2\sqrt{3}+7i+2\sqrt{3}}{7}=\frac{7i}{7}=i$

5 0

3 years ago

Find the average value of f over region

SVETLANKA909090 [29]

$D$ is a right triangle with base length 1 and height 8, so the area of $D$ is $\dfrac12(1)(8)=4$ .

The average value of $f(x,y)$ over $D$ is given by the ratio

$\dfrac{\displaystyle\iint_Df(x,y)\,\mathrm dA}{\displaystyle\iint_D\mathrm dA}$

The denominator is just the area of $D$ , which we already know. The average value is then simplified to

$\displaystyle\frac74\iint_Dxy\,\mathrm dA$

In the $x,y$ -plane, we can describe the region $D$ as all points $(x,y)$ that lie between the lines $y=0$ and $y=8x$ (the lines which coincide with the triangle's base and hypotenuse, respectively), taking $0\le x\le1$ . So, the integral is given by, and evaluates to,

$\displaystyle\frac74\int_{x=0}^{x=1}\int_{y=0}^{y=8x}xy\,\mathrm dy\,\mathrm dx=\frac78\int_{x=0}^{x=1}xy^2\bigg|_{y=0}^{y=8x}\,\mathrm dx$
$=\displaystyle56\int_{x=0}^{x=1}x^3\,\mathrm dx$
$=14x^4\bigg|_{x=0}^{x=1}$
$=14$

3 0

2 years ago