Hey there!!
Fill in the blanks :-
⇒ First graph the line. Locate the <u>value of x </u>on the x-axis. Draw a vertical line from <u>point plotted on the x-axis </u>to the graph of the function and a horizontal line segment from the graph of the function to the y-axis.
<em>Find the value of f(x) when x is -2. </em>
Remember :- <u><em>f(x) is basically the y-value. It is just denoted as _f(x), it stands for function of x. Which means, the value of y, depends upon the value of x or the function of x. </em></u>
Given : x = - 2
Plugging in the values :
...
...
...
The last fill in the blank :
The value of y on the y-axis is the value of the function. Therefore, the value of f(x) is <u>-7 </u>when x is -2.
Hope it helps!!
For Sammie the equation would be S=3N.
If Nora (N) has 4 goldfish, you substitute N=4 into the previous equation to find out how many goldfish Sammie has.
> S= 3 x 4
= 12
The next part of the question is asking you how many goldfish do Sammie and Nora have in all. If Sammie has 12 goldfish and Nora has 4 goldfish, the equation would be 12 + 4= 16
The answer would be 16 goldfish in total.
The answer is: 10
Hope this helps! (:
Since no value is given to Eva, we'll call her x. Justin, then, is 7.50+x. That leaves Emma, who would be Justin (7.50+x) -12. If we add them all together we get 63. So our equation is
x + (7.50+x) + (7.50+x)-12 = 63
Combine like terms
x+x+x + 7.50+7.50-12 = 63
3x + 3 = 63
3x = 60
x = 20
Since we said Eva is x then Eva has 20 dollars.
Justin is 7.50 + x so 7.50+20=
$27.50
Emma is Justin - 12 so 27.50-12= $15.50
Let's check our answer...
15.50+27.50+20 = 63
63=63
It checks!
Problem 1
<h3>Answer:
6.7</h3>
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Work Shown:
The two points are and
Apply the distance formula to get the following
The distance between the two endpoints is roughly 6.7 units. This is the same as saying the segment is roughly 6.7 units long.
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Problem 2
<h3>Answer: 3.6</h3>
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Work Shown:
We'll use the distance formula here as well.
This time we have the two points and
The distance between them is...
This distance is approximate.