Can someone please help me on this please I beg you no links I really appreciate it plz

Please help with the top 3 i’ll give u brainliest, what is the volume of the figures below?

jasenka [17]

Answer:

im sorry but i only know the second one, i believe its 224.3

Step-by-step explanation:

8 0

3 years ago

Does this table represent a function?why or why not

Alex787 [66]

The answer is C, because there are two different numbers correlated to the same number on the Y side. The table does not represent a function.

6 0

3 years ago

Read 2 more answers

Phoebe took a survey of her classmates' favorite sport. The results are in the table below:

olga55 [171]

Answer:

.14

Step-by-step explanation:

7 0

3 years ago

A liquid shake is being served in a cone shaped container with a base diameter of 8 centimeters and a height of 22 centimeters.

zhuklara [117]

Answer:

The container weighs approximately 28 g.

Step-by-step explanation:

In order to solve this problem we first need to calculate the volume of the container, since it's shaped like a cone we must use the following formula:

volume container = (pi*r²*h)/3

Applying the data from the problem:

r = d / 2 = 8/2 = 4 cm

volume container = [pi*(4)²*22]/3

volume container = (pi*16*22)/3

volume container = (352*pi)/3 = 368.6135 cm³

Since the density is 0.82 g/cm³, we can find the mass of the shake by using the following formula:

density = mass / volume

shake mass = density*volume = 0.82*368.6135 = 302.26307 g

Since the container plus shake weighs 330 grams, then the container alone weighs:

container = 330 -302.26307 = 27.73693 g

The container weighs approximately 28 g.

5 0

3 years ago

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