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alukav5142 [94]

3 years ago

13

Solve by Completing the Square

1 answer:

guajiro [1.7K]3 years ago

4 0

You have to do -b/2a and put a square at the end of each =. You have to make it equal to 0 first though

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The graph below plots a function f(x): graph of line segment going through ordered pairs 0, 100 and 3, 220 If x represents time,

Alja [10]

m = (y2 - y1)/(x2 - x1)

m =(220 - 100)/(3 - 0)

m = 120/3

m = 40

Hope this helps!

~LENA~

5 0

3 years ago

Read 2 more answers

The first triangle is dilated to form the second triangle.

KATRIN_1 [288]

Hello, the correct answers are

truth for 0.25

false for 4

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8 0

3 years ago

What are the zeros of the function f(x)=x2+12x+38

Artemon [7]

Answer:

The zero would be -2.714 (x-intercept)

6 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

What two numbers add to be 6 and multiply to be 15

yarga [219]

$\displaystyle \bf \\
x_1+x_2=6\\\\
x_1\cdot x_2=15\\\\
\Longrightarrow~~x^2-6x+15=0\\\\
x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{36-60}}{2}=\\\\
=\frac{6\pm\sqrt{-24}}{2}=\frac{6\pm2\sqrt{-6} }{2}=3\pm i\sqrt{6}\\\\
x_1=\boxed{3-i\sqrt{6}}\\\\
x_2=\boxed{3+i\sqrt{6}}\\\\
\texttt{Verify:}\\\\
x_1+x_2=3-\underline{i\sqrt{6}}+3+\underline{i\sqrt{6}}=3+3=6~~correct\\\\
x_1\cdot x_2=(3-i\sqrt{6})(3+i\sqrt{6})=3^2-(i\sqrt{6})^2=9+6=15~~correct$

4 0

4 years ago