Five and three sixths minus two and one third

Round 363 to the nearest ten and hundred

IRISSAK [1]

Its 400 because 363 is closer to 400 than 300

3 0

4 years ago

Read 2 more answers

Hurry please find the values of x and y

MrRissso [65]

Answer:
Y=70
X= 20
You do first the adjacent angles 180-40 so 140 then u have that the triangle is isosceles so the bas angles are equal 180 -140 = 40
40 divided by 2 20 and 20 so x= 20
Then 90 -20 is 70
So 180 - (70+40)
180-110=70 so y= 70
Hope you understand

8 0

3 years ago

Beth is knitting mittens and gloves. Each pair must be processed on three machines. Each pair of mittens requires 2 minutes on M

anygoal [31]

Answer:

Two mittens and seven gloves

Step-by-step explanation:

I presume you are asking how many mittens and how many gloves Beth should knit to produce the most profit.

Let x = the number of mittens and

Let y = the number of gloves

1. Summarize the information in a table

$\begin{array}{cccc}&\textbf{Mittens}& \textbf{ Gloves} & \textbf{Minutes Available}\\\text{Machine A} & 2 & 4 & 32\\\text{Machine B} & 2 & 2 & 18\\\text{Machine C} & 4 & 1 & 24\\\text{Unit profit} & $8.00 & $10.00 & \\\end{array}$

2. List the constraints

(a) Beth can't make a negative number of mittens, so

x ≥ 0.

(b) Neither can she make a negative number of gloves, so

y ≥ 0.

(c) The total number of minutes on Machine A cannot exceed 32, so

2x + 4y ≤ 32.

(d) The total number of minutes on Machine B cannot exceed 18, so

2x + 2y ≤ 18.

(e) The total number of minutes on Machine C cannot exceed 24, so

4x + y ≤ 24.

3. Write an equation for the profit

P = 8.00x + 10.00y

4. Graph the constraints to identify the feasible region

See the figure below. The feasible region is the dark green area where all three regions overlap.

5. Determine the points of intersection between the constraints

(i) Constraints (c) and (d)

$\begin{array}{rcr}2x + 4y & = & 32\\2x + 2y & = & 18\\2y & = & 14\\y & = & \mathbf{7}\\2x + 14 & = & 18\\2x & = &4\\x & = & \mathbf{2}\\\end{array}\\$

The lines intersect at (2,7).

(ii) Constraints (c) and (e)

$\begin{array}{rcl}2x + 4y & = & 32\\4x + y & = & 24\\4x + 8y & = & 64\\7y & = & 40\\y & = & \mathbf{\dfrac{40}{7}}\\\\4x + \dfrac{40}{7} & = & 24\\\\4x & = &24 - \dfrac{40}{7}\\\\4x & = & \dfrac{128}{7}\\\\x & = & \mathbf{\dfrac{32}{7}}\\\\\end{array}\\$

The lines intersect at (27/7,32/7).

(iii) Constraints (d) and (e)

$\begin{array}{rcr}2x + 2y & = & 18\\4x + y & = & 24\\4x + 4y & = & 36\\3y & = & 12\\y & = & \mathbf{4}\\4x +4 & = & 24\\4x & = &20\\x & = & \mathbf{5}\\\end{array}\\$