Answer:
c. 2a3+9a2+45a+6ab2+18b2
Step-by-step explanation:
(a+3) + (−2a2+15a+6b2)
Using distributed p. with and a and 3
a(−2a2) + a(15a) +a(6b2) and 3(−2a2) + 3(15a) + 3(6b2)
= -−2a3 + 15a2 + 6ab2 = -6a2 + 45a + 18b2
Now add like terms and put in standard form.
15a2+(-6a2) = 9a2
And equals
-2a3 + 9a2+45a+6ab2+18b2
Trust me I took the test and got 100%
Your welcome,
Alright, so we know that the race is 5 kilometers, so the equation will be 5-<some value>= <distance from finish line>. We also know that the student runs a kilometer every three minutes, so 3x=1km . Multiplying both sides by 5, we get 15y=5km (y being the number to make the equation make sense, or the slope). When the student has run 5km, the distance from the student to the finish line should be 0, so we get that 5-5=0, and plugging 15y in for 5 we get 5-15y=0. For 15x to equal 5, 3y=1 and y=1/3. Therefore, we plug that in for y, getting 5-15(1/3)=0. However, we have to make it for all times! Since 15 represents the minutes, we make that x, and since 0 represents the distance remaining, we make that the distance remaining, making it 5-(1/3)x=distance left. You can also think of y as the slope in y=mx+b - it stays constant that way.
-230 because it is farther away from 0
-2,-3
-3, -4
-4, -5
you are showing the line of growth in the graph.
