A graph is shown below:
1 answer:
You might be interested in
1. 
2. -2+5n or 5n-2
3. (12s)+150 where s=number of students
4. 3(32)+1(53)=96+53=159 per hour, so 159h
9. √105 is irrational, -4 is an integer, 4/3 is a rational number, order is -4,3/4, √105
10. 3²=9, 4²=16, so between 3 and 4, but closer to 4, so √14≈4
11. that is the associative property, it deals with moving parentahees around
12. 2(6.25)+1(5.50)+2(2.75)=2(6.25+2.75)+1(5.50)=2(9)+5.5=18+5.5=23.5
13. yes, because they are both 2 less than 3 times of a number
14. maybe 18 times 3, so (4²+2) times 3=54
2. -2+5n or 5n-2
3. (12s)+150 where s=number of students
4. 3(32)+1(53)=96+53=159 per hour, so 159h
9. √105 is irrational, -4 is an integer, 4/3 is a rational number, order is -4,3/4, √105
10. 3²=9, 4²=16, so between 3 and 4, but closer to 4, so √14≈4
11. that is the associative property, it deals with moving parentahees around
12. 2(6.25)+1(5.50)+2(2.75)=2(6.25+2.75)+1(5.50)=2(9)+5.5=18+5.5=23.5
13. yes, because they are both 2 less than 3 times of a number
14. maybe 18 times 3, so (4²+2) times 3=54
Answer: 
Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.
Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.
Now, since we know the end behaviours, let's find the asymptotic behaviours.
As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.
So, our range would be:
Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.
Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.
Now, since we know the end behaviours, let's find the asymptotic behaviours.
As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.
So, our range would be:
Answer:
The mass of the lamina is 1
Step-by-step explanation:
Let be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:
.
From the question, the given density function is .
Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
The mass of the lamina can be found by evaluating the double integral:
.
Since D is a rectangular region, we can apply Fubini's Theorem to get:
.
Let the inner integral be: , then
.
The inner integral is evaluated using integration by parts.
Let , the partial derivative of u wrt y is
and
, integrating wrt y, we obtain
Recall the integration by parts formula:
This implies that:
We substitute the limits of integration and evaluate to get:
This implies that:
.
Or
.
We again apply integration by parts formula to get:
.
.
.
.
No unit is given, therefore the mass of the lamina is 1.