All categories

3 years ago

Ignore this i don’t know how to delete jt

Mathematics

2 answers:

saveliy_v [14]3 years ago

8 0

Answer:

Step-by-step explanation:

pshichka [43]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t an

Alenkasestr [34]

Answer: 693 years

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

$A(t) =A_0e^{kt}$

k = rate constant

t = time taken for decomposition = 1

$A_0$ = Initial amount of the reactant

$A_t$ = amount of the reactant left = $A_0-\frac{0.1}{100}\times A_0=0.999A_0$

$0.999A_0=A_0e^{k\times 1}$

$0.999=e^k$

$k=-0.001year^{-1}$

for half life : $t=t_\frac{1}{2}$

$A_t=\frac{1}{2}A_o$

Putting in the values , we get

$\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}$

$t_\frac{1}{2}=693 years$

Thus half life of this isotope, to the nearest year is 693.

8 0

4 years ago

I am not understanding these problems. Help me, please?

yanalaym [24]

Answer:

I think I know the answer but not 100% sure

Step-by-step explanation:

Step 1: Complete the scatter plot

Step 2: After you get done, submit the answer and see what grade you get.

7 0

3 years ago

If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of a?

Harman [31]

Answer:

......................................,........

8 0

3 years ago

Please help Divide. 9u2− 12u/3u

podryga [215]

It has to be at lest 9u2-4?

3 0

3 years ago

write an equation of the perpendicular bisector of the line segment whose endpoints are (-1,1) and (7,-5)

gladu [14]

Find midpoint of segment.
(-1,1), (7,-5)
$x= \frac{-1+7}{2}=3,y= \frac{1+(-5)}{2}=-2$

Find slope of line which passes through (-1,1) and (7,-5).
$m= \frac{-5-1}{7-(-1)}= \frac{-6}{8}=- \frac{3}{4}$

The slope of perpendicular line is negative reciprocal value of m.
$m_{\perp}=- \frac{1}{m} =- \frac{1}{- \frac{3}{4} } = \frac{4}{3}$

Write quation of line which slope is $\frac{4}{3}$ and passes throug the point (3,-2).
$m=\frac{4}{3},x_1=3,y_1=-2
\\y-y_1=m(x-x_1)
\\
\\y-(-2)=\frac{4}{3}(x-3)
\\
\\y+2=\frac{4}{3}x-\frac{8}{3}
\\
\\3y+6=4x-8
\\4x-3y-14=0$

3 0

3 years ago