Answer:
The missing statement is ∠ACB ≅ ∠ECD
Step-by-step explanation:
Given two lines segment AC and BD bisect each other at C.
We have to prove that ΔACB ≅ ΔECD
In triangle ACB and ECD
AC=CE (Given)
BC=CD (Given)
Now to prove above two triangles congruent we need one more side or angle
so, as seen in options the angle ∠ACB ≅ ∠ECD due to vertically opposite angles
hence, the missing statement is ∠ACB ≅ ∠ECD
Answer:

Step-by-step explanation:
Hello,
Based on the indication, we can write this polynomial as below, k being a real number that we will have to identify (degree = 3 and we have three zeroes -3, -1, and 2).

We know that the point (1,10) is on the graph of this function, so we can say.

Then the solution is:

Hope this helps.
Do not hesitate if you need further explanation.
Thank you
Answer:
the answer is A
Step-by-step explanation: because the way you work out problems with parenthesis is Parenthesis Exponents Multiplication Division Adding Subtraction and what ever id in the parenthesis comes first
Answer:
([-3], [0]), ([3], [0])
Step-by-step explanation:
The given equation of the hyperbola is presented as follows;

The vertices of an hyperbola (of the form)
are (± a, 0)
The given hyperbola can we presented in a similar form as follows;

Therefore, by comparison, the vertices of the parabola are (± 3, 0), which gives;
The vertices of the parabola are ([-3], [0]), ([3], [0]).
Step-by-step explanation:
Let us consider the task to find the angle between vectors ES and EJ (the first letters are taken to name the vectors).
\overrightarrow{ES} = (4;4) - (4; -3) = \overrightarrow{(0; 7)}
ES
=(4;4)−(4;−3)=
(0;7)
\overrightarrow{EJ} = (-5; -4) - (4; -3) = \overrightarrow{(-9; -1)}
EJ
=(−5;−4)−(4;−3)=
(−9;−1)
cos \alpha=\frac{\overrightarrow{ES}*\overrightarrow{EJ}}{|\overrightarrow{EJ}|*|\overrightarrow{ES}|}cosα=
∣
EJ
∣∗∣
ES
∣
ES
∗
EJ
cos(a) = (0*(-9)+7*(-1)) / (7*9.055) = -0.11043;
a = 96,34°
Solution: 96 degrees.