Answer:
2c^2 - 2c - 3
Step-by-step explanation:
Next time, please share the answer choices. Thank you.
Group together the 'c' terms in 2 + 7c – 4c2 – 3c + 4:
2 + 4c and -4c^2 + 4 Then we have:
-4c^2 + 4c + 6
We should reduce this by factoring -2 out of all three terms:
2c^2 - 2c - 3 This is the desired quadratic in standard form.
Answer:
(3, 50) and (14,303)
Step-by-step explanation:
Given the system of equations;
y=23x–19 ....1
x²–y= – 6x–23 ...2
Substitute 1 into 2;
x²–(23x-19)= – 6x–23
x²–23x+19= – 6x–23 .
x²-23x + 6x + 19 + 23 = 0
x² - 17x + 42 = 0
Factorize;
x² - 14x - 3x + 42 = 0
x(x-14)-3(x-14) = 0
(x-3)(x-14) = 0
x = 3 and 14
If x = 3
y = 23(3) - 19
y = 69-19
y = 50
If x = 14
y = 23(14) - 19
y = 322-19
y = 303
Hence the coordinate solutions are (3, 50) and (14,303)
Both of those equations are solved for y. So if the first one is equal to y and the second one is equal to y, then the transitive property says that the first one is equal to the second one. We set them equal to each other and solve for x. 4x-5=-3 and 4x = 2. That means that x = 1/2. We were already told that y = -3, so the coordinates for the solution to that system are (1/2, -3), B from above.
Answer:
3
Step-by-step explanation:
if the shapes are similar then the measures are proportional:
4/12 = 1/d cross multiply expressions
4d = 12 divide both sides by 4
d = 3