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3 years ago

Which of the following represents

Mathematics

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

The correct option is (c).

Step-by-step explanation:

The given number is -0.904.

We need to represent the number in scientific notation.

Any number can be written in scientific notation as follows :

$a\times 10^b$

Where

a is a real number and b is an integer

We can write it as follows :

$-0.904=-9.04\times 10^{-1}$

Hence, the correct option is (c).

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114 pages of a novel which was 3 pages less than 1/3 
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3 years ago

F(x)=2x^2 -5 , simplify f(x+h)/h , where h≠0

7nadin3 [17]

Answer:

$\huge\boxed{\dfrac{f(x+h)}{h}=\dfrac{2x^2+4xh+2h^2-5}{h}}$

$\huge\boxed{\dfrac{f(x+h)}{h}=\dfrac{2x^2-5}{h}+4x+2h}$

Step-by-step explanation:

$f(x)=2x^2-5\\\\\dfrac{f(x+h)}{h}=\dfrac{2(x+h)^2-5}{h}=\dfrac{2(x^2+2xh+h^2)-5}{h}\\\\\dfrac{f(x+h)}{h}=\dfrac{2x^2+4xh+2h^2-5}{h}\\\\h\neq0$

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3 years ago

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

3 years ago