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3 years ago

I NEED HELP ASAP!!PLEASE HELP ME WITH MY MATH!

Mathematics

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer:

A. -2, 9

B. -2, 4

C. -8, 4

D. -8, 9

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Two numbers have a difference of 17 and a sum of 23 which system of equations can be used to determine x and y the two numbers?

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3 years ago

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

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Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$ $A=45e^{-11.25}$ $A=45\times0.000013=0.000585$ $A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams

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1 year ago

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A projectile is fired at 571 mi hr. How fast is that in meters per second to the nearest whole value? Use the conversion factor

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Answer:

Consider this: factors that are the same that are in the numerator and denominator of a fraction cancel out.

e.g. 20/30 = [2(10)]/[3(10)] = 2/3 This is because the 10 present in the numerator exactly equals the 10 present in the denominator.

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