The answer is
5mn(KUASA DUA) -6gh(KUASA DUA) +2
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer:
Step-by-step explanation:
To compare 3,176 and 3,472 using the number line, remember how numbers appear on a number line:
Smaller numbers are to the left, and larger numbers are to the right.
but since both is in the 3,000 just make sure one is on the left and find out where the bigger one goes on the right
The given function is
A . f(4) means we have to evaluate the function when x = 4.
<h2>Therefore, f(4) = 14.</h2>
B. f(x) = 50 means we have to evaluate the function when y = 50.-
<h2>Therefore, f(16) = 50.</h2>
C. f(-5) means we have to evaluate the function when x = -5.
<h2>Therefore, f(-5) = -13.</h2>
D. f(x) = -34.
<h2>Therefore f(-12) = -34.</h2>
Add the following polynomials:
All polynomials are already ordered according to their degree (exponent), and therefore you can just add straight down in columns.
For x3: 2 + 1 - 3 = 0
For x2: -4 + 6 + 2 = 4
For x: 6 - 8 -4 = -6
Constants: -3 + 12 - 7 = 2
The answer is:
4x^2 - 6x + 2
