Answer: B
Step-by-step explanation:
Answer:
coefficient of x^2 is -4
and degree of polynomial is 3
With the given information, we can create several equations:
120 = 12x + 2y
150 = 10x + 10y
With x being the number of rose bushes, and y being the number of gardenias.
To find the values of the variables, we can use two methods: Substitution or Elimination
For this case, let us use elimination. To begin, let's be clear that we are going to be adding these equations together. Therefore, in order to get the value of one variable, we must cancel one of them out - it could be x or y, it doesn't matter which one you decide to cancel out. Let's cancel the x out:
We first need to multiply the equations by numbers that would cause the x's to cancel out - and this can be done as follows:
-10(120 = 12x + 2y)
12(150 = 10x + 10y) => Notice how one of these is negative
Multiply out:
-1200 = -120x - 20y
+ 1800 = 120x + 120y => Add these two equations together
---------------------------------
600 = 100y
Now we can solve for y:
y = 6
With this value of y known, we can then pick an equation from the beginning of the question, and plug y in to solve for x:
120 = 12x + 2y => 120 = 12x + 2(6)
Now we can solve for x:
120 = 12x + 12 => 108 = 12x
x = 9
So now we know that x = 9, and y = 6.
With rose bushes being x, we now know that the cost of 1 rose bush is $9.
With gardenias being y, we now know that the cost of 1 gardenia is $6.
Answer:
y = 18 and x = -2
Step-by-step explanation:
y = x^2+bx+c To find the turning point, or vertex, of this parabola, we need to work out the values of the coefficients b and c. We are given two different solutions of the equation. First, (2, 0). Second, (0, -14). So we have a value (-14) for c. We can substitute that into our first equation to find b. We can now plug in our values for b and c into the equation to get its standard form. To find the vertex, we can convert this equation to vertex form by completing the square. Thus, the vertex is (4.5, –6.25). We can confirm the solution graphically Plugging in (2,0) :
y=x2+bx+c
0=(2)^2+b(2)+c
y=4+2b+c
-2b=4+c
b=-2+2c
Plugging in (0,−14) :
y=x2+bx+c
−14=(0)2+b(0)+c
−16=0+b+c
b=16−c
Now that we have two equations isolated for b , we can simply use substitution and solve for c . y=x2+bx+c 16 + 2 = y y = 18 and x = -2
A because (-5)^2 = 25 + 2(5) =35-1=34
