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3 years ago

First 2 common multiples of 8 and 9

Mathematics

2 answers:

Alik [6]3 years ago

7 0

The first two common multiples of 8 and 9 are 72 and 144!

RSB [31]3 years ago

3 0

8:0,8
9:0,9

Hope this helped

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Collin is making a bulletin board and note center. He is using square cork tiles and square dry-erase tiles. One of every 3 squa

Lisa [10]

Well 12 divided by 3 is 4 so there will be 4 cork squares.

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3 years ago

What is the slope of the line shown below?

Artyom0805 [142]

Answer:

Step-by-step explanation:

Use the coordinates of the given points to find the slope.

difference in y-coordinates = -3 - 6 = -9

difference in x-coordinates = 2 - (-1) = 3

slope = -9/3 = -3

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3 years ago

Stella graphs the equation y=1/3x – 2

faltersainse [42]

we have

$y=\frac{1}{3}x-2$

Statements

case A) The graph is a straight line.

The statement is True

Because, this is a linear equation (see the attached figure)

case B) The line passes through the origin.

The statement is False

Because the point $(0,0)$ is not a solution of the equation

Verify

Substitute the value of x and y in the equation

$0=\frac{1}{3}*0-2$

$0=-2$ ------> is not true

the point $(0,0)$ is not a solution

therefore

The line does not pass through the origin

case C) The line passes through the point $(0,-2)$

The statement is True

Because the point $(0,-2)$ is a solution of the equation

Verify

Substitute the value of x and y in the equation

$-2=\frac{1}{3}*0-2$

$-2=-2$ ------> is true

the point $(0,-2)$ is a solution

therefore

The line passes through the point $(0,-2)$

case D) The slope of the line is $3$

The statement is False

Because, the slope of the line is $m=\frac{1}{3}$

case E) The y-intercept of the line is $2$

The statement is False

we know that

The y-intercept is the value of y when the value of x is equal to zero

For $x=0$

find the value of y

$y=\frac{1}{3}*0-2=-2$

the y-intercept is equal to $-2$

5 0

3 years ago

Read 2 more answers

226Ra has a half-life of 1599 years. How much is left after 1000 years if the initial amount was 10 g?

Ray Of Light [21]

$\bf \textit{Amount for Exponential Decay using Half-Life}\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &10\\
t=\textit{elapsed time}\to &1000\\
h=\textit{half-life}\to &1599
\end{cases}
\\\\\\
A=10\left( \frac{1}{2} \right)^{\frac{1000}{1599}}$