The second cylinder is taller correct?
8 in < 11 in
If the height is also larger, the base is most likely larger too.
Wouldn't the largest cylinder have a Larger surface area?
20x^4-39x^3-16x
Make sure to multiply all the terms by each other in the two parentheses
Answer:
c= 12t
c is the amount of energy you burn, so it'll be the "solution" to the equation. Multiply 12 times the amount of time you spend exercising to find how much energy you'll burn.
Answer: 20
H(c) = 6.4 + 0.6c
<u>6.4</u> is the constant.
When the height of the cups is <u>18.4</u> the function is:
18.4 = 6.4 + 0.6c
Then, you add <u>6.4</u> from both sides
18.4 - 6.4 + 6.4 = 6.4 + 0.6c - 6.4 + 6.4
Simplify
18.4 = 6.4 + 0.6c
Switch sides
6.4 + 0.6c = 18.4
Multiply both sides by <u>10</u>
6.4 x 10 + 0.6c x 10 = 18.4 x 10
Refine
64 + 6c = 184
Subtract <u>64</u> from both sides
64 + 6c - 64 = 184 - 64
Simplify
6c = 120
Divide both sides by <u>6</u>
6c/6 = 120/6
c = <u>20</u>
Answer:
<h2>see below</h2>
Step-by-step explanation:
<h3>
Question-6:</h3>
we are given a equation

to solve so
recall logarithm multiplication law:

simplify multiplication:

remember 
so

cancel out
from both sides:

simplify squares:

move left hand side expression to right hand side and change its sign:
since we are moving left hand side expression to right hand side there'll be only 0 left in the left hand side

rewrite it to standard form i.e ax²+bx+c=0

rewrite -6x as 2x-8x:

factor out x and 8:

group:



<h3>Question-7:</h3>
move left hand side log to right hand side:

use mutilation logarithm rule;

so

cancel out log from both sides:

make it standard form:

factor:

so
