SA=2(lw+wh+lh) This is the formula for finding the surface area of a rectangular prism, where SA is surface area, l is length, w is width, and h is height.
208=2(lw+wh+lh)
104=lw+wh+lh Here, I divided both sides by 2 to get ride of the 2.
Now, I used prime factorization to find out all the prime factors of 104, which are 2, 2, 2, and 13. Since rectangular prisms only have 3 dimensions, I needed to combine two of the prime factors. In this case, I can either combine 2 of the 2s to get 2, 4, and 13 or I can combine 13 with one of the 2s to get 26, 2, and 2.
If my dimensions were 2, 4, and 13...
my surface area would be 172 sq cm.
If my dimensions were 2, 2, and 26...
my surface area would be 208 sq cm.
Hence, the width of the rectangular prism when the surface area is 208 square centimeters can be either 2 or 26.
Answer:
11,8,5
Step-by-step explanation:
an=11-3(n-1)
Let n=1
a1 = 11 - 3(1-1)
= 11 -0
=11
Let n=2
a2 = 11-3(2-1)
= 11 -3(1)
= 8
Let n=3
a3 = 11-3(3-1)
= 11 -3(2)
= 11 -6
=5
Answer:
Below
Step-by-step explanation:
First we can go ahead and create a general equation for this polynomial
Here are our roots :
x1 = - 3
x2 = -1
x3 = 1
Now because this function extends from quadrant 4 to 3, we know that this has been reflected in the x-axis :
f(x) = - ( x + 3 ) ( x + 1 ) ( x - 1 )
However if we look closely you can see that the graph appears to "bounce" off certain roots. In this case it bounces off x = 1. This means that this root is an order of 2. It also has a weird looking curve on x = - 3 which means that this root is an order of 3.
Our general equation will look like this :
f(x) = - ( x + 3 )^3 ( x - 1 )^2 ( x + 1 )
Now we need to sub in any point on the graph to solve for the <em>a </em>value. I'm just going to arbitrarily pick the y-intercept at ( 0 , -3 )
- 3 = - a ( 0 + 3 )^3 ( 0 - 1 )^2 ( 0 + 1 )
- 3 = - a (3)^3 (-1)^2 (1)
- 3 = - a (27)(1)(1)
- 3 = - a27
1/9 = a
Here is our FINAL equation :
f(x) = - 1/9 ( x + 3 )^3 ( x - 1 )^2 ( x + 1 )
Hope this helps! Best of luck <3
I would really appreciate a brainliest if possible :)
