Answer:
1.
C. 6561 is the answer that the table represents
Answer: x = 24
Step-by-step explanation:
Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.
<u>Step 1:</u><u> </u><u><em>Remove parentheses.</em></u>
12 (2 − 2) + 32 =
+ 6 + 2
<u>Step 2:</u><u> </u><u><em>Simplify 2 − 2 to 0.</em></u>
12 × 0 + 32 =
+ 6 + 2
<u>Step 3:</u><u> </u><u><em>Simplify 12 × 0 to 0.</em></u>
0 + 32 =
+ 6 + 2
<u>Step 4:</u><u> </u><u><em>Simplify 0 + 32 to 32.</em></u><em> </em>
32 =
+ 6 + 2
<u>Step 5:</u><u> </u>
<u><em> + 6 + 2 to </em></u>
<u><em> + 8.</em></u>
32 =
+ 8
<u>Step 6:</u><u> Subtract 8 from both sides.</u>
32 − 8 =
<u>Step 7:</u><u> Simplify 32 − 8 to 24.</u>
24 =
<u>Step 8:</u><u> </u><u><em>Switch sides.</em></u>
= 24
~I hope I helped you! :)~
Adding 14.52 and -14.52 cancels it out
The answer is 6.89
I hope that helps!
It depends on the value of the discriminant of the quadratic.
If you are familiar with the quadratic formula...
x=(-b±√(b^2-4ac))/(2a)
The part under the radical sign (b^2-4ac) is called the discriminant of the quadratic (of the form ax^2+bx+c)
If the discriminant is less than zero there are no REAL solutions (however there are two imaginary solutions)
If the discriminant is equal to zero, there is just one solution
If the discriminant is greater than zero, there are two real solutions.
So depending on context, if you are to exclude imaginary solutions (which we often do) you can have 0,1, or 2 real solutions.
However you will NEVER have three solutions. (real or imaginary)
Formula is
ₙCₖ = (n!)/( k!*(n-k)! )
so ₇C₃ = (7!) / (3! * (7-3)!)
and 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1
= (7 * 6 * 5 * 4 * 3 * 2 * 1)/(3 * 2 * 1 * 4!)
= (7 * 6 * 5 * 4)/(4!)
= (7 * 6 * 5 * 4)/(4 * 3 * 2 * 1)
= (7 * 6 * 5 )/(6 * 1)
= (7 * 5 )/( 1)
= 35