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3 years ago

Find A ∩ B given A = {the Integers} and B = {the Rational numbers}.

Mathematics

1 answer:

garri49 [273]3 years ago

8 0

Step-by-step explanation:

Since, every integer can be expressed as rational number.

$\therefore A\cap B = B \\$

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

professor190 [17]

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

4 0

1 year ago

jeka57 [31]

Answer:

Step-by-step explanation:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\(-5,1)\quad\implies\quad x_1=-5\,,\ \ y_1=1\\\\(-1,4)\quad\implies\quad x_2=-1\,,\ \ y_2=4\\\\d=\sqrt{(-1+5)^2+(4-1)^2}= \sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

4 0

3 years ago

Need help for this question asap please

mr_godi [17]

Answer:

option 4 = 25

Step-by-step explanation:

given= chord js and hL intersects at poimt p

to find = value of x

solution:according to the therom;

if the line extensions of the chordJS and HL INTERSECTS at point K then their lenghts satisfy

JK×KS=HK×KL

10×30=X×12

300=12x

300/12=x

25= x

3 0

3 years ago

A movie rental company offers a membership plan that costs $10/month and charges $2 to rent each movie. How many movies can a me

Brums [2.3K]

Answer:

See below

Step-by-step explanation:

linear equation: y=2x+10

x=cost per movie rented

10=$10 for the membership

24=2x+10 (24 being total you want to spend)

subtract 10 from each side

14=2x

divide by 2

7=x

He can rent no more than 7 movies if he want to spend $24 or less in the month

8 0

3 years ago

Which expression has the same GCF as 15x2 - 21x?

Mice21 [21]

Answer:

$15x^2 +21 x$

Step-by-step explanation:

We have the following expression:

$15x^2 -21 x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are -21,-7,-3,-1,1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

If we consider the new expression:

$15x^2 +21x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are 1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

And as we can see both GCF are equal. So then we satisfy the condition required.

5 0

3 years ago

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