51 feet you have to multiple 32 by 8 it will give you 256 and then divide 256 and 5 and you get 51
Answer:
P(F/C) = 0.75
P(C^F) = 0.09
P(C^F) = P(F^C)
Step-by-step explanation:
P(C^F) = 9% = 0.09
P(C^F) = P(F^C)
P(F/C) = P(F^C)/P(C)
= 0.09/0.12 = 0.75
First, we add the numbers. This gives us 715.
Then, we divide it by the amount of numbers. This gives us. 715 divided by 11.
This simplifies to 65.
Hi there!
The question here asks us to find all the roots, or solutions, of the quartic equation x⁴ - 4x³ + x² + 12x - 12 = 0. We can start off by using the Rational Root Theorem, which calls for the constant over the leading coefficient, which for us would be 12 over 1. Since we need to list all the factors of 12, they all become + or -.
<u>+</u>12 / <u>+</u> 1
=> <u>+</u>1, <u>+</u>2, <u>+</u>3, <u>+</u>4, <u>+</u>6, <u>+</u>12 / <u>+</u>1
Now, just simplify each one. We get the possible roots 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12. The next part is the most tedious. Plug in each possible factor for x and see if the resulting value is 0. We get that only root 2 makes the whole equation equal to zero. However, we are not done yet. The question asks for the ALL the roots of the equation, not just real, but also imaginary. Let's use synthetic division to find the other roots.
We can set the divisor to 2, as (x - 2) = 0 is equal to x = 2.
2| 1 -4 1 12 -12
2 -4 -6 12
1 -2 -3 6 0
We can now factor the whole original equation to (x - 2)(x³ - 2x² -3x + 6) = 0. Now, we have to factor the expression x³ - 2x² -3x + 6, and grouping seems like the most appropriate way.
x³ - 2x² -3x + 6 = 0
(x³ - 2x²) + (-3x + 6) = 0
x²(x - 2) - 3(x - 2) = 0
(x² - 3)(x - 2) = 0
Next, use the Zero Product Property to get the values of x.
x² - 3 = 0
x² = 3
x = √3
x ≈ <u>+</u> 1.73
x - 2 = 0
x = 2
Therefore, the roots of the equation x⁴ - 4x³ + x² + 12x - 12 = 0 are x = 2,x = 2 (repeated solution), x = 1.73, x = -1.73. Hope this helped and have a great day!
The equation for resolving couples will come in place

Using the unit vectors by i and j along the x-axis and the y-axis you'll express it as a Cartesian vector.
<h3>
Couple moment and Cartesian vector</h3>
Couple moment:This looks at the product of the forces acting on a body( in this context a steering wheel ) by the perpendicular distance that exist amidst the forces.
Cartesian vector:this is the representation of unit vectors by i and j along the x-axis and the y-axis, respectively.
Therefore, in calculating the <em>couple</em> moment t acting on the steering wheel, the equation for resolving couples will come in place

Using the unit vectors by i and j along the x-axis and the y-axis you'll express it as a Cartesian vector.
More on Cartesian Co-ordinate
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