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3 years ago

Solve the equation by completing the square

Mathematics

1 answer:

vovikov84 [41]3 years ago

4 0

You solve by useing the formula (b/2) ^2
And your answer is x= 3,1

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Harold wrote this equation to model the level of water in a pool over time. The variable x represents time in hours. f(x) = 3,50

lina2011 [118]

Answer:

The water level is falling.

The initial level of water in the pool was 3,500 units

The water was 2,600 units high after 4 hours.

Step-by-step explanation:

The given function that models the water level is

$f(x)=3,500-225x$

where $x$ represents time in hours.

The function represents a straight line that has slope $m=-225$

Since the slope is negative, it means the water level is falling.

The initial level of water in the pool can found when we put $x=0$ into the function.

$f(0)=3,500-225(0)$

$f(0)=3,500$ , hence the initial level is 3,500.

To determine the level of water in the pool after 14 hours, we put $x=14$ into the equation to get;

$f(14)=3,500-225(14)$

$f(14)=3,500-3150$

$f(14)=350$

To determine the water level after 4 hours we put $x=4$

$f(4)=3,500-225(4)$

$f(4)=3,500-900$

$f(14)=2,600$

4 0

3 years ago

Read 2 more answers

You picked 8 flowers and your friend picked 17

Ann [662]

Answer:

Your friend picked 9 more flowers

Step-by-step explanation:

Take the amount your friend picked and subtract the amount you picked

17 - 8

Your friend picked 9 more flowers

6 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

4 years ago

You have 59 total coins for a total of 12..05. You only have quarters and dimes. How many of each coin do you have?

vovikov84 [41]

41 quarters and 18 dimes.

6 0

3 years ago

Read 2 more answers

The base of the pyramid is a square of side 12cm.

atroni [7]

Answer:

384 .cm2