Hello!
<u>Number 22
</u>
: We'd plot the first point at 0 since there is no stated y-intercept. Next, we'd use our slope to determine where to plot the next point, and that would create our line. According to the problem, our slope is
, which automatically tells us that the slope would be going downwards because it's negative.
To plot our point, use the slope while going down and across from our y-intercept, which is 0. Go down 1, and over 2.
Your points should be at (0, 0) and (-1, 2)
<u /><u>Number 23:</u> This one will be a bit trickier since the equation is not in slope-intercept form. First, let's convert it to slope-intercept form.
Flip some of those numbers around to get our equation in slope-intercept form:
Now to graph this, we do the same as we did for the last problem. Plot our first point at (0, 2), since 2 is our y-intercept. Afterwards, go up 2 and over 3, then plot the other point.
Your points should be at (0, 2) and (4, 3)
To solve for this, let's try the highest number x and y can be.
X < 5, so it can be 4.
Y < 6, so it can be 5.
4 + 5 = 9.
9 < 11, so x + y < 11.
Your missing inequality symbol is:
<
I hope this helps!
We have the following number:
23.5
This number can be written as:
23 + 0.5
Or:
23 + 1/2
So, the new expression gives us two new numbers. In this way, we can say that:
23 = Twenty three
1/2 = a half
In conclusion, the number 23.5 can be written as follows:
<em>23.5 = Twenty three and a half</em>
<h3><u>The expression gives the number of transistors in a dense integrated circuit in 1979 is:</u></h3>
<em><u>Solution:</u></em>
<em><u>The increasing function is given as:</u></em>
Where,
y is the future value
a is the initial value
r is the growth rate
t is the number of years
From given,
a = 5000
t = 1974 to 1979 = 5 years
<em><u>Substituting the values in formula,</u></em>
Thus the expression gives the number of transistors in a dense integrated circuit in 1979 is found