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3 years ago

Simplify. remove grouping sbols and combine like terms

Mathematics

1 answer:

harina [27]3 years ago

3 0

Answer:

$5a^2+12b^2-40b$

Step-by-step explanation:

First distribute the -4 to the 2b and -3b. This leaves you with $5a^2-8b+12b(b+4)$

Now you can distribute the 12b to b and +4. This leaves you with $5a^2-8b+12b^2+48b$

Now you can combine your like terms leaving you with $5a^2+12b^2-40b$ .

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Diego went out for a walk. He walked for 20 minutes East at a rate of 5km/hour. He then turned around and walked west for 10 min

iVinArrow [24]

Positive 20 and negative 10 is a net amount of 10. 10 minutes is 1/6 of an hour. 5*1/6=5/6km

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3 years ago

Im bored in this quarentine and have a lot of work to catch up on. One of my math problems is "How much water should be added to

Kazeer [188]

Answer:

20/3 ounces of water is added.

Step-by-step explanation:

how many ounces of pure water should be added to 20 ounces of 40% solution of muriatic acid to obtain a 30% solution of muriatic acid?

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

The acid is 40% of 20 ounces = 8 ounces.

0.4*20 = 8

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That same 8 ounces is to be 30%. The water will be the other 70%.

8/30 = x/100

30x = 800

x = 80/3, the total amount of the solution.

We started with 20, so 80/3 - 20 = 20/3

20/3 ounces of water is added.

8 0

3 years ago

How do I find the integral<br> ∫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

60 (x + 20) 3x what is the value of x?

Whitepunk [10]

Answer:

Step-by-step explanation:

Step 1: Equation at the end of step 1

Simplifying

60(x + 20) = 3x

Reorder the terms:

60(20 + x) = 3x

(20 * 60 + x * 60) = 3x

(1200 + 60x) = 3x

Solving

1200 + 60x = 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.

1200 + 60x + -3x = 3x + -3x

Combine like terms: 60x + -3x = 57x

1200 + 57x = 3x + -3x

Combine like terms: 3x + -3x = 0

1200 + 57x = 0

Add '-1200' to each side of the equation.

1200 + -1200 + 57x = 0 + -1200

Combine like terms: 1200 + -1200 = 0

0 + 57x = 0 + -1200

57x = 0 + -1200

Combine like terms: 0 + -1200 = -1200

57x = -1200

Divide each side by '57'.

x = -21.05263158

Simplifying

x = -21.05263158

7 0

2 years ago

Write y= -0.5x+0.8 in standard form using integers

pickupchik [31]

Standard form is Ax+By=C

0.5x+y=0.8

8 0

3 years ago

Read 2 more answers