12 subtracted from a number equals 25 find the number

First, "boxes of two sizes" means we can assign variables: Let x = number of large boxes y = number of small boxes "There are 115 boxes in all" means x + y = 115 [eq1] Now, the pounds for each kind of box is: (pounds per box)*(number of boxes) So, pounds for large boxes + pounds for small boxes = 4125 pounds "the truck is carrying a total of 4125 pounds in boxes" (50)*(x) + (25)*(y) = 4125 [eq2] It is important to find two equations so we can solve for two variables. Solve for one of the variables in eq1 then replace (substitute) the expression for that variable in eq2. Let's solve for x: x = 115 - y [from eq1] 50(115-y) + 25y = 4125 [from eq2] 5750 - 50y + 25y = 4125 [distribute] 5750 - 25y = 4125 -25y = -1625 y = 65 [divide both sides by (-25)] There are 65 small boxes. Put that value into either equation (now, which is easier?) to solve for x: x = 115 - y x = 115 - 65 x = 50 There are 50 large boxes.

3 0

3 years ago

Maria throws a softball straight up into the air with an initial velocity of 50 ft/sec. The ball leaves her hand when it is exac

Nataly_w [17]

Answer:

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

Step-by-step explanation:

Let suppose that effects from air friction and Earth's rotation can be neglected, so that the softball can be modelled experiment a free fall, that is, an uniform accelerated motion due to gravity. From we know the initial velocity and position of the position and we can determine the final position of the ball as a function of the final velocity:

$v_{f}^{2} = v_{o}^{2}+2\cdot a \cdot (s_{f}-s_{o})$ (1)

Where:

$s_{o}$ , $s_{f}$ - Initial and final position of the softball, measured in feet.

$a$ - Acceleration, measured in meters per square second.

$v_{o}$ , $v_{f}$ - Initial and final velocities of the softball, measured in feet per second.

If we know that $v_{o} = 50\,\frac{ft}{s}$ , $a = -32.174\,\frac{m}{s^{2}}$ and $s_{o} = 4\,ft$ , then the equation that models the situation is:

$v_{f}^{2} = 2500-64.348\cdot (s_{f}-4)$

Then, we clear the final position of the softball:

$-\frac{v_{f}^{2}-2500}{-64.348} = s_{f}-4$

$s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

8 0

2 years ago

If the order of three loci is A B C, and the map distance between A and B is 15 m.u., and the map distance between B and C is 20

zysi [14]

Answer:

Therefore, we conclute that the map distance between A and C is 35 m.u..

Step-by-step explanation:

We know that the order of three loci is A B C, and the map distance between A and B is 15 m.u., and the map distance between B and C is 20 m.u. We calculate the map distance between A and C.

Therefore, we get

$AC=AB+BC\\\\AC=15+20\\\\AC=35\\$

Therefore, we conclute that the map distance between A and C is 35 m.u..

4 0

3 years ago