my class collected 389 milk jugs. We used 40 jugs. We gave 30 milk to the first graders. how many milk jugs did we have?

1 answer:

8 0

We will just do the following steps:-

389 - 40 = 349
349 - 30 = 319

So, in the end your class had 319 milk jugs left.

Hope I helped ya!!

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Hi! im alex, just answer these easy questions and recive your points <3

Delvig [45]

Answer

106 sio so you multiply first

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2 years ago

Consider the polynomial function g(x)=x³-3x²+6x+8. Divide g by x+1.

irina [24]

Answer:

Quotient ( x² -4x +10)

Reminder -2

Step-by-step explanation:

We have to consider the polynomial function g(x) = x³ - 3x² + 6x + 8

Now, we have to divide g(x) by (x+1)

Let us arrange the terms of g(x) to get (x+1) as common.

x³ - 3x² + 6x + 8

= x³ +x² -4x² -4x +10x +10 -2

= x² (x +1) -4x (x +1) +10 (x +1) -2

= (x +1)( x² -4x +10) -2

Hence, if we divide g(x) by (x +1) then the quotient will be ( x² -4x +10) and the reminder will be -2. (Answer)

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2 years ago

Find a formula for the described function. A rectangle has perimeter 16 m. Express the area A of the rectangle as a function of

Alex17521 [72]

Part A:

Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.

Let the length of the other side be x, then

$2(L+x)=16 \\ \\ \Rightarrow L+x=8 \\ \\ \Rightarrow x=8-L$

Thus, if the length of one of the side is x, the length of the other side is 8 - L.

Hence, the area of the rectangle in terms of L is given by

$L(8 - L) = 8L-L^2$

Part B:

To find the domain of A

Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.

Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.

Hence the domain of the area are the set of values between 0 and 8 not inclusive.

Therefore,

$dom(Area)=0\ \textless \ L\ \textless \ 8 \ or \ (0, 8)$