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SOVA2 [1]
3 years ago
6

The sum of two consecutive odd integers is − 68 -68 . Find the numbers.

Mathematics
1 answer:
algol133 years ago
8 0

Answer:

Step-by-step explanation:

let one no. be 'x'.

the other no. is (x+2)

therefore the eq. is:

x+x+2= -68

2x+2= -68

2x = -68-2

    = -70

x= -70/2

 = -35

one number is -35

other number is (x+2) = -35 + 2

                                     = -33

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The hypotenuse of a right triangle is 14 centimeters long. One of the legs of the triangle is 6 centimeters. What is the length
Fed [463]
Use Pythagorean theorem to solve.
a^2 + b^2 = c^2
6^2 + b ^2 = 14^2
36 + b^2 = 196
Subtract 36 from both sides.
b^2 = 196-36
b^2 = 160
Take the square root of both sides.
b = sqrt 160
As a decimal
b = 12.649
As a simplified radical
b = 4sqrt10
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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
Given log5 3=0.6826 and log5 8=1.2920, evaluate the expressions<br><br> Please help!?
gulaghasi [49]

Answer:

a) 1.3652

b) 1.3906

Step-by-step explanation:

a) log5 9 = log5 3² = 2log5 3 = 2(0.6826) = 1.3652

b) log5 75/8 = log5 75 - log5 8 = log5 3×25 - log5 8=

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3 0
2 years ago
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A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
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