For example, on Google you may see that "1.5 cubic foot boxes are the standard box, manufactured by most companies". So, we assume that a standard cardboard box is 1.5 feet tall.

So, if we set the equation

$1.5k = 5 \iff k=\dfrac{5}{1.5}=3.\bar{3}$

So, there would be between 3 and 4 cardboard boxes in a 5 feet tall stack.

Similarly, we can see that the average book is 9 inches tall. 9 inches are 0.75

feet, so we have

$0.75k=5 \iff k=6.\bar{6}$

So, there would be between 6 and 7 books in a 5 feet tall stack.

The average brick is 75 millimeters tall, which means 0.25 feet tall. Again, we have

$0.25k=5 \iff k=20$

So, there would be 20 bricks in a 5 feet tall stack.

Finally, a coin is about 0.006 feet, which leads to

$0.006k=5 \iff k=833.\bar{3}$

So, there would be between 833 and 834 coins in a 5 feet tall stack.

5 0

3 years ago

#4 need help with this one

4vir4ik [10]

Answer:

x = 25.6

Step-by-step explanation:

As we can see, this is a right triangle, so there are known relationships

well to start we have to know the relationship between angles, legs and the hypotenuse

α = 43

a: adjacent = x

o: opposite

h: hypotenuse = 35

sin α = o/h

cos α = a/h

tan α = o/a

we see that it has (angle, adjacent, hypotenuse)

cos α = a/h

cos 43° = x / 35

cos 43° * 35 = x

25.59 = x

if we want to round to the nearest tenth

x = 25.59 = 25.6

4 0

3 years ago

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