What is 60 increased<br> by 20%

A delivery truck is transporting boxes of two sizes: large and small. The combined weight of a large box and a small box is 85

insens350 [35]

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

Why does the Easter bunny carry eggs? Rabbits don't lay eggs.

Sonja [21]

Answer:

He is hungry.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0

2 years ago

Write an equation to match each graph:

Makovka662 [10]

Answer:

y=-|x+1|+1

Step-by-step explanation:

It's an absolute value function vertically flipped and shifted 1 unit to the left, 1 unit to the the top.

8 0

3 years ago

I don’t get this answer

matrenka [14]

23 units sure does hope this helps

3 0

3 years ago

What is 60 increased by 20%