Answer:
-60
Step-by-step explanation:
-3 x -4 x -5
12 x -5
-60
negative x negative x negative= negative
Answer:
9/2 if n goes to infinity and that the 2n^3 is under the whole expression
Step-by-step explanation:
Let me clear this .
find limit (9n^3 + 5*n - 2)/ (2n^3)
as n --> infinity
Did I put the parentheses in the right spot?
because if you leave it the way you did, then the whole expression goes to positive infinity as n goes to infinity But I will do this with parentheses
so
find limit (9n^3 + 5*n - 2)/ (2n^3)
simplify expression
limit (9/2) + 5/(2n^2) - 1/(n^3)
= (9/2) + 0 - 0
= (9/2)
A = 900
The area of a parallelogram can be found by taking its base (b) and multiplying by the height (h). The base has the measure of 33 +12 = 45 cm. Now, you just need to find the height of the parallelogram. The height can be found by using the pythagorean theorem (

.
So, when simplified the height of the parallelogram will be found from

The height is 20 cm. The base is 45 cm. 45 x 20 = 900.
Answer:
There is not a slope perpendicular to the lie x = -6
Step-by-step explanation:
In this equation, you are given a line, x = -6.
This is in fact a line with an undefined slope. The slope for a perpendicular intersection is the opposite reciprocal of the slope.
Eg. If you have the line y = 2x
The reciprocal is y = 1/2x, the opposite means negative
So the slope that would intersect this line perpendicularly is y = -1/2x.
There is an undefined slope in this, therefore, you cannot find the slope for a line perpendicular to x = -6.
If it were an option, I would choose "no slope"
Answer:
<h2>The perimeter of the cross section is 30 centimeters.</h2>
Step-by-step explanation:
In this problem we have the intersection of a rectangular prism an a plane.
The dimensions of the rectangular plane are

Assuming the plane is cutting the prism horizontally, the cross section would have dimensions

Because only the height would be cut.
So, the perimeter of the rectangle cross-section is

Therefore, the perimeter of the cross section is 30 centimeters.