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2 years ago

14

The midpoint of AB is M(0,2). If the

1 answer:

otez555 [7]2 years ago

8 0

Answer:

Step-by-step explanation:

(x+2)/2 = 0

x + 2 = 0

x = -2

(y - 3)/2 = 2

y - 3 = 4

y = 7

(-2, 7)

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HELP ASAP!!! IM BEING TIMED. Use synthetic division to test one potential root. Enter the numbers that complete the division pro

PtichkaEL [24]

Step-by-step explanation:

1 * (-5) = -5 a = -5

6 + (-5) = 1 b = 1

1 * (-5 ) = -5 c = -5

-7 + -5 = -12 d = -12

8 0

2 years ago

Read 2 more answers

The 3 side lengths of a right-

Gekata [30.6K]

Answer: the area of the triangle = 10.5 cm²

Step-by-step explanation:

Let 3x=a, 4x=b, 5x=c.

Hence,

P=2*(a+b+c)

36=2*(3x+4x+5x)

36=2*12x

36=24x

Divide both parts of the equation by 24:

1.5 cm=x

a=3x

a=3*1.5

a=4.5 cm

b=4x

b=4*1.5

b=6 cm

c=5x

c=5*1.5

c=7.5 cm

$\displaystyle\\S=\frac{a*b}{2} \\\\S=\frac{4.5*6}{2} \\\\S=3.5*3\\\\S=10.5\ cm^2$

3 0

1 year 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

A cone has a volume of 12 cubic inches. What is the volume of a cylinder that the cone fits exactly inside of?

Stells [14]

36in^3
as the volume of a cone is 1/3 that of the volume of a cylinder the smallest cylinder that the cone can fit inside of is one that is exactly 3x its the volume

5 0

3 years ago

Read 2 more answers

Four graphs are shown below:

Aleks [24]

graph b because only a and b are linear associations and only b and c are negative

5 0

2 years ago