Refer to the image attached.
Given:
and
are congruent.
To Prove:
ABC is an isosceles triangle.
Construction: Construct a perpendicular bisector from point B to Line segment AC.
Consider triangle ABD and BDC,
(given)
(By the definition of a perpendicular bisector)
(By the definition of a perpendicular bisector)
Therefore,
by Angle Side Angle(ASA) Postulate.
Line segment AB is congruent to Line segment BC because corresponding parts of congruent triangles are congruent.(CPCTC)
Answer:
Number 1
Step-by-step explanation:
The median in number 1 is 13 and the range for number 1 is also thirteen so it has to be it.
Answer:
(If they were talking about 3,335, then) 100 times greater
(If they were talking about 3,335 then) 10 times greater
Step-by-step explanation:
When you visualize it, ignore the other numbers and think of only the threes.
Count how many spaces/places they are from each other, and that's how many 0s you should put after a 1 as your answer.
Same as decimal moving and multiplying by 10's.
Y = -2x + 4
substitute the value of x = 1 to the equation:
y = -2 · 1 + 4 = -2 + 4 = 2
Answer: a. 2
Answer:
Step-by-step explanation:
Let 
Subbing in:

a = 9, b = -2, c = -7
The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

Group together in groups of 2:

Now factor out what's common within each set of parenthesis:

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

Remember that 
so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring
gives us that x = 1 and -1. The other set is a bit more tricky. If
then
and

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:
±
which will simplify down to
±
Those are the 4 solutions to the quartic equation.