4(5 - 3)^3 - 2(6 - 4)^3
We can simplify the expression by applying the order of operations, which states that we must first simplify any parentheses, then exponents, then perform multiplication or division from left to right, and finally, addition or subtraction from left to right.
First, we simplify within each set of parentheses:
4(5 - 3)^3 - 2(6 - 4)^3
4(2)^3 - 2(2)^3
Next, we simplify the exponents:
4(8) - 2(8)
Then, we perform the multiplication:
32 - 16
Finally, we subtract:
32 - 16 = 16
Therefore, the result is 16.
2(-5)^2 - 3(-3)^2
We can simplify the expression by applying the order of operations.
First, let's simplify within each set of parentheses:
2(-5)^2 - 3(-3)^2
2(25) - 3(9)
Next, we perform the exponents:
2(25) - 3(9)
Then, we perform the multiplication:
50 - 27
Finally, we subtract:
50 - 27 = 23
Therefore, the result is 23.
-39\sqrt{5} simplified
To simplify -39√5, we can first rewrite it as -39 * √5 since the negative sign in front can be factored out.
Next, we can evaluate the square root:
√5 ≈ 2.236
Therefore, -39 * √5 is approximately equal to -87.204.
\sqrt{2}\left(\sqrt{10}-6\sqrt{5}\right)
To simplify √2(√10 - 6√5), we can apply the distributive property.
We start by multiplying the √2 with each term inside the parentheses:
√2 * √10 - √2 * 6√5
√2 * √10 simplifies to √(2 * 10) = √20 = √(2 * 2 * 5) = 2√5.
Next, we multiply √2 by 6√5:
√2 * 6√5 = 6 * √(2 * 5) = 6 * √10.
Putting it all together, we have:
2√5 - 6√10
This is the simplified form of the expression √2(√10 - 6√5).
The maximum number of volts, E, that can be placed across a resistor is given by the formula E = square root of (PR), where P is the number of watts of power that the resistor can absorb and R is the resistance of the resistor in ohms. Find E if P = (1/4) watts and R = 3600 ohms.
Question 8 options:
To find E, we substitute the given values of P and R into the formula E = √(PR).
Given:
P = 1/4 watts
R = 3600 ohms
Substituting these values, we have:
E = √((1/4) * 3600)
Simplifying the expression within the square root:
E = √(900)
Taking the square root of 900, we get:
E = 30
Therefore, the maximum number of volts, E, that can be placed across the resistor is 30 volts.