The <em>correct answers</em> are:
A) Both can be solved by graphing;
C) Both can be solved by substitution; and
D) Both have solutions at the points of intersection.
Explanation:
Just as a system of linear equations can be solved by graphing, a system of quadratic equations can as well. We graph both equations. We then look for the intersection points of the graphs; these intersection points will be the solutions to the system.
We can also solve the system by substitution. If we can get one variable isolated, we can substitute this into the other equation to solve.
Answer:
3.2537496e+30
Step-by-step explanation:
Answer:
16
Step-by-step explanation:
We can see that Ar is the perpendicular bisector of chord BD. Since A is the center of the circle, AR is the radius of the circle, which is 10 (6+4)
Next, we can see that when we connect point A to point D, it is also a radius. Thus, AD is also equal to 10 as the radius of the circle remains the same.
Using Pythagoras theorem, a^2 + b^2 = c^2, we can make a right angled triangle of ACD.
AC = 6 = a
CD = ? = b
AD = 10 = c
10^2 = 6^2 + b^2
b^2 = 10^2 - 6^2 = 64
b = CD = 8
Now, since Ar is the perpendicular bisector of chord BD, BD = CD x 2
BD = 8 x 2 = <u>16</u>
Answer:
All you need to remember is the rules
Step-by-step explanation:
Let us remember
a to the m power x a to the nth power is = a to the m+n power. (add the exponents)
And
a to the m power ÷ a to the nth power is = a to the m-n power. (subtract the exponents) So
14 to the -4 power x 14 to the 7 power= 14 to the -4+7 which is equal to 14 to the 3rd power
Answer:
111.625
Step-by-step explanation:
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