Select the correct answer.

Bess [88]

It's forming a circle so we know that together it's going to add up to 360. #1 is 66 because angle A trough E is straight and we have 90 on angle ACF and it's vertical to angle FCG. #2 is 125 because angle ACB is 24 because it's vertical to angle GCE. #3 is 114 because 90 and 24 is 114. #4 is 156 because 66 and 90 equal 156.
1. FCG = 66
2. BCD = 125
3. FCB = 114
4. ACG = 156

4 0

4 years ago

7) David uses 2 1/4 cups of dough to make a

Dafna11 [192]

Answer:

9 3/8 ÷ 2 1/4

Step-by-step explanation:

too lazy too do it, plus if ur having a hard time doing fractions on a calc, just do 1÷4 as a replacement for .25 when shown 1/4. just go to frac calc in the future it makes it easy ro use....

7 0

3 years ago

an art student wants to take a triangle with side lengths of 9,9, and 15 and enlarge it so that the longest side of a new triang

VikaD [51]

Answer:

13.5

Step-by-step explanation:

The longest side of the smaller triangle is 15 while that of the bigger one is 22.5

The enlargement ratio is 22.5/15 = 1.5

so the lengths of the smaller sides will be;

(1.5*9) = 13.5

6 0

3 years ago

Suppose that 66% of a large sample of a population favor increasing the speed limit. If there are 160,000 people in the populati

Step2247 [10]

B is the answer i think

3 0

3 years ago

Read 2 more answers

Suppose a certain manufacturing company produces connecting rods for 4- and 6-cylinder automobile engines using the same product

gregori [183]

Answer:

Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder to W_4 or 0 is

$x_4 \le W_4 * s_4$

$x_4 \le 5000 * s_4$

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder to W_6 or 0 is

$x_6 \le W_6 * s_6$

$x_6 \le 8,000 * s_6$

Generally the constrain that limits the production of connecting rods for both 4 cylinder and 6 cylinders is

$x_4 \le W_4 * s_6$

=> $x_4 \le 5000 * s_6$

$x_4 \le W_6 * s_4$

=> $x_4 \le 8000 * s_4$

$s_4 + s_6 = 1$

The minimum cost of production for next week is

$U = M_4 * x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6$

=> $U = 13x_4 + 16x_6 + 2000 s_4 + 3500 s_6$

Step-by-step explanation:

The cost for the four cylinder production line is $C_4 = \$2,100$

The cost for the six cylinder production line is $C_6 = \$3,500$

The manufacturing cost for each four cylinder is $M_4= \$13$

The manufacturing cost for each six cylinder is $M_6= \$16$

The weekly production capacity for 4 cylinder connecting rod is $W_4 = 5,000$

The weekly production capacity for 6 cylinder connecting rod is $W_6 = 8,000$

Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder to W_4 or 0 is

$x_4 \le W_4 * s_4$

$x_4 \le 5000 * s_4$

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder to W_6 or 0 is

$x_6 \le W_6 * s_6$

$x_6 \le 8,000 * s_6$

Generally the constrain that limits the production of connecting rods for both 4 cylinder and 6 cylinders is

$x_4 \le W_4 * s_6$

=> $x_4 \le 5000 * s_6$

$x_4 \le W_6 * s_4$

=> $x_4 \le 8000 * s_4$

$s_4 + s_6 = 1$

The minimum cost of production for next week is

$U = M_4 * x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6$

=> $U = 13x_4 + 16x_6 + 2000 s_4 + 3500 s_6$

3 0

3 years ago