I already a seed you when you asked this the first time. The answer is 3.
You multiply both sides by 50 and then you get this
(In the picture)
Answer:
392
Step-by-step explanation:
Answer:
Step-by-step explanation:
vertex form of quadratic: y = a ( x − h ) ^ 2 + k
h= -2 , k = -2
y=a(x+2)^2-2
Substitute (-1,1)
1=a(-1+2)^2-2
a=3
y=3(x+2)^2-2
Expanding and give
y=3x^2+12x+10
Answer:
Let's begin by writing the mathematical equations given by the sentences.
Let X = smaller number
Let Y = larger number
The first sentence tells us
X + Y = 38
The second sentence tells us
X = Y/3 + 6
Since you weren't given a specific method to use, you can use either the substitution method or the elimination(addition) method, whichever works best for you.
Since the 2nd equation has the X isolated, I'm going to isolate the X in the first equation,
X = 38 - Y
Since both equations are equal to X, I can set them equal to one another and solve for Y.
Y/3 + 6 = 38 - Y
Y + 18 = 114 - 3Y
4Y = 96
Y = 24
Then use either of the two original equations to solve for X
X + Y = 38
X + 24 = 38
X = 14
Step-by-step explanation:
PLS MAKE ME AS BRAINLIST
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
