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Ostrovityanka [42]
2 years ago
14

-1/2x=-10

= - 10 " alt=" - 1 \div 2x = - 10 " align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Rainbow [258]2 years ago
6 0



x= 1/20 is the answer


Triss [41]2 years ago
5 0

Answer:

x= 1/20

Step-by-step explanation:

-1/2x = -10

multiply by 2x on each side

2x(-1/2x) = -10*2x

-1 = -20x

divide by -20

-1/-20 = -20x/-20

1/20 =x

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If a radioactive isotope has a half-life of 6 hours and 12 milligrams is initially present, how much will
Serhud [2]

Answer:0.75 milligrams will  be present in 24 hours

Step-by-step explanation:

Step 1

The formula for radioactive decay can be written as

N(t)=No (1/2)^(t/t 1/2)

where

No= The amount of the radioactive substance at time=0=milligrams

t 1/2= the half-life= 6 hours

t=24 hours

N(t) = the amount at time t

Step 2--- Solving

N(t)=No (1/2)^(t/t 1/2)

=N(t)=12 x ( 1/2) ^ (24/6)

= 12 x (1/2) ^4

= 12 x 0.0625

= 0.75 milligrams

3 0
2 years ago
Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ
atroni [7]

Answer: y=Ce^(^3^t^{^9}^)

Step-by-step explanation:

Beginning with the first differential equation:

\frac{dy}{dt} =27t^8y

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

\frac{1}{y} \frac{dy}{dt} =27t^8

Multiply both sides by 'dt' to get:

\frac{1}{y}dy =27t^8dt

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt

ln(y)=27(\frac{1}{9} t^9)+C

ln(y)=3t^9+C

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)

y=e^(^3^t^{^9} ^+^C^)

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

y=e^(^3^t^{^9}^)e^C

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

y=Ce^(^3^t^{^9}^)

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8

Now check if the derivative equals the right side of the original differential equation:

(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)

Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0
2 years ago
Please round to the nearest tenth if possible.
Butoxors [25]
\bf \cfrac{64~\underline{mi}}{\underline{hr}}\cdot \cfrac{\underline{hr}}{60~sec}\cdot \cfrac{5200~ft}{\underline{mi}}\implies \cfrac{64\cdot 5200~ft}{60~sec}\implies \cfrac{16640~ft}{3~sec}
\\\\\\
\approx 5546.67\frac{ft}{sec}\\\\
-------------------------------\\\\
\cfrac{6000~\underline{lbs}}{\underline{day}}\cdot \cfrac{ton}{2000~\underline{lbs}}\cdot \cfrac{7~\underline{day}}{week}\implies \cfrac{6000\cdot 7~ton}{2000~week}\implies 21.00\frac{ton}{week}

\bf -------------------------------\\\\
\cfrac{2~\underline{lbs}}{\underline{week}}\cdot \cfrac{16~oz}{\underline{lbs}}\cdot \cfrac{\underline{week}}{7~day}\implies \cfrac{2\cdot 16~oz}{7~day}\implies \cfrac{32~oz}{7~day}\approx 4.57\frac{oz}{day}
8 0
3 years ago
Help please!!<br> asap no wrong answers<br> will give the crown symbol
Svetllana [295]

Answer:

4Hz

Step-by-step explanation:

Standard form of a sine or cosine function,

y = acos(b(x+c))

where a is the amplitude, b is the value to find the period. and c is the phase shift.

Period = \frac{2\pi}{b}

From the equation given in the question,

y = 3cos(8\pi \: t +  \frac{\pi}{2} ) \\ y = 3cos(8\pi(t +  \frac{1}{16} )) \:  \: (factorising \: 8\pi \: out)

We can see:

Amplitude = 3,

Period = \frac{2\pi}{8\pi} = 1 / 4

Phase Shift = 1 / 16

Now we want to find the frequency.

Frequency = 1 / Period

= 1 / (1/4)

= 4Hz

4 0
2 years ago
Read 2 more answers
Samuel has 52 marbles in a bag. Fourteen are red, 13 are green and 25 are blue. If a marble is chosen at random, what is the pro
lorasvet [3.4K]
.25
you can come to this conclusion by calculating 13/52
7 0
2 years ago
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