Can anyone please help me?

Please show your work simplify the expression (-5-c)(-1)

frozen [14]

(-5-c)(-1)
can write as (-1)(-5-c)
multiply(-1) to (-5) then multiply (-1) to(-5)
answer 5+c

4 0

3 years ago

In 2000 South Dakota had a population of 754,844. The number of people in South Dakota increased by 7.9% in 2010. What is South

Novay_Z [31]

814,180. Hope I helped. If i didn't I am sos sorry. I would take 7.9 or .079 to 754,844 to get your number.

6 0

3 years ago

Find the Perimeter & Area for the following circle.

Feliz [49]

Answer:

A = 28.3 cubic inches; P = 20.1 inches

Step-by-step explanation:

Looks like 3/4 of a full circle

The area of a circle is calculated with the formula: $\pi r^{2}$

Since the problem wants pi = 3.14, we use 3.14 instead of pi in the equation.

The radius is given as 3 inches

Area = $3.14*3^{2} = 3.14*9 = 28.26$ cubic inches = 28.3 cubic inches

------------------------

Perimeter is calculated with $2\pi r$ .

But this is 3/4 of a circle with an additional two 3-inch sides

Perimeter = $\frac{3}{4}*2\pi r$ + 2(3) = 14.1 + 6 = 20.1 inches

6 0

2 years ago

Read 2 more answers

Help help help! 25 points

Paraphin [41]

Step-by-step explanation:

-6x^2 -8x +2x-8 = -6x^2 -6x -8

it's a quadratic.

8 0

2 years ago

Read 2 more answers

How do I find the integral ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago