Slope intercept form is y = mx + b
so make the equation look like that.
-9x + 10y = -9 ... add 9x on both sides
10y = -9 + 9x
10y = 9x - 9
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y = (9/10)x - (9/10)
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114/6=9 parts per minute. 9 parts * 15 minutes= 135 parts
Answer:
(a)
A to B to C =40 yards
(b)
w<40
Step-by-step explanation:
We are given
The distance between point A and point B is 15 yards
AB=15 yards
The distance between point B and point C is 25 yards
BC=25 yards
(a)
A to B to C = AB+BC
A to B to C =15+25
A to B to C =40 yards
(b)
direct distance between A to C or AC
It must be less than distance between A to B to C
AC< AB+BC
AC< 15+25
AC< 40 yards
w<40
I think it might be B
because from 9+1 = 10 we can change to 9=10-1
and 10+6=16 no need to change
then we solve the problem with the equation
16 - 9
10+6 - (10-1)
10+6-10+1
7
thats the answer
If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial. C is our constant, 2, and d is the leading coefficient, 1. The factors of 2 are +/- 1 and +/-2. The factors for 1 are +/-1 only. Meaning, in all, there are 4 possibilities as roots for this polynomial. But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above. Let's try x = -1, factor form (x + 1). If there is no remainder when we do the synthetic division, then -1 is a root. Put -1 outside the "box" and the coefficients from the polynomial inside: -1 (1 2 -1 -2). Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1. Put that -1 up under the 2 and add to get 1. Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2. -1 times -2 is 2, and -2 + 2 = 0. So we have our first root of (x+1). The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!). The new polynomial is

. That can also be factored to find the remaining 2 roots. Use standard factoring to find that the other 2 solutions are (x+2) and (x-1). Our solutions then are x = -2, -1, 1, choice B from above.