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3 years ago

Cobalt-60 is used in the treatment of cancer. A sample contains 8 grams of Co-60 where k = 0.1308

Mathematics

2 answers:

masha68 [24]3 years ago

8 0

You need to post more information.

coldgirl [10]3 years ago

5 0

Answer:

Step-by-step explanation:

5.3

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Which expression is equivalent to b^m x b^n? A. b^m+n B. b^m-n C. b^mx n D. b^m/n

noname [10]

The formula for multiplying exponents are such below.

(b^m)^n = b^mn
b^m/b^n=b^(m-n)
b^m x b^n=b^(m+n)

4 0

3 years ago

Read 2 more answers

What is the the correct order for least to greatesed in these fractions,5/6,7/12 and 3/4

scZoUnD [109]

Answer:

7/12 3/4 5/6

Step-by-step explanation:

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3 years ago

If P(AnB)=2/3 and P(B)=3/4, what is P(A I B)? A.35/36 B.15/16 C.24/25 D.8/9

astraxan [27]

Answer:

$\large\boxed{D.\ \dfrac{8}{9}}$

Step-by-step explanation:

$\text{We know:}\\\\P(A|B)=\dfrac{P(A\ \cap\ B)}{P(B)}\\\\\text{We have:}\\\\P(A\ \cap\ B)=\dfrac{2}{3},\ P(B)=\dfrac{3}{4}.\\\\\text{Substitute:}\\\\P(A|B)=\dfrac{\frac{2}{3}}{\frac{3}{4}}=\dfrac{2}{3}:\dfrac{3}{4}=\dfrac{2}{3}\cdot\dfrac{4}{3}=\dfrac{8}{9}$

6 0

3 years ago

They’re 80 skittles in a bag 20% of them are read how many skittles a red

Nutka1998 [239]

Answer: 16

Step-by-step explanation: 80x20=1600

Put the decimal place where it belongs

80% of 20=16

Hope this helps!

7 0

3 years ago

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

8 0

3 years ago