Solve for w. 5w + 9z = 2z + 3w w = z w = z w = –2z w = –7z
1 answer:
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You need a system of 3 equations to solve for the 3 unknowns in our parabola, standard form of 
. Since the second point there has an x of 0, we'll start there to make it easy. In this point, x = 0 and y = -4. 
. That gives us our first value...c = -4. Let's do the first equation now the same way, but this time we have a c value to sub in: 
and 
which simplifies to 
. We'll use that in a bit. Let's do the third point now the same way. 
and 
which simplifies to 
. Now we have a new system of equations. We need to solve for a and b. Let's multiply the 2nd equation by -4 to get rid of the a terms. Doing that we have 
and we will add that to . The a terms cancel each other out leaving us with 48=12b and b = 4. Now we'll sub that b into one of the equations in terms of a and b and solve for a. -16 = 16a + 4(4) and -16=16a+16. Subtracting 16 from both sides and we have -32=16a and a = -2. Here's what we have for our values now: a = -2 b = 4, c = -4. So the quadratic in standard form is 
. And you're done!
Answer:
m<DEF = 8y - 6
Step-by-step explanation:
Given that: DEG = , and FEG =
.
The given triangle is an isosceles triangle with sides DE ≅ DF.
But,
m<DEF = DEG + FEG
= +
= 3y + 4 + 5y - 10
= 8y - 6
m<DEF = 8y - 6
Therefore, the measure of < DEF is 8y - 6.