For f(x)=1/x^2-3
Find
A) f(3)
B) f(2-h)
If f(x)=1/x^2-3, then f(3) = 1 / 3^2 - 3. The exponentiation here must be carried out first: f(3) = 1/9 - 3. Then f(3) = 1/9 - 27/9 = -26/9
If f(x)=1/x^2-3, then f(2-h) = 1 / [2-h]^2 - 3. This result may be left as is or expanded. In expanded form, we have:
1
f(2-h) = ------------------ - 3
4-4h +h^2
The length of AQ is 4, the length of WQ is 12, 4 divided by 12 can be simplified as 1 divided by 3 so the answer is A. 1/3
<h3>Option A</h3><h3>3x - y = -5 is the equation of the line with a slope 3 and y intercept 5</h3>
<em><u>Solution:</u></em>
<em><u>The slope intercept form of equation is given as:</u></em>
y = mx + c ------ eqn 1
Where,
m is the slope of line
c is the y intercept
From given,
slope = m = 3
y intercept = c = 5
<em><u>Substitute m = 3 and c = 5 in eqn 1</u></em>
y = 3x + 5
3x - y = -5
Thus the equation of line is found
Answer:
It is a function.
Step-by-step explanation:
Functions are relations in which one domain value is assigned to exactly one range value. The x-value must not repeat in the relation set for it to be a function.
{(2,3), (4,3), (6,3), (5,9), (3,9)}
There are no repeating x-values given.
The relation should be a function.
Hope this helps.
The answer is c 10 #s are greater than 110
