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emmainna [20.7K]
3 years ago
10

A small box measures 3 1/3 in. by 2 3/4 in. by 1 5/6 in. high. What is the volume of the box? Show your work

Mathematics
1 answer:
Mashutka [201]3 years ago
7 0

Answer:

The volume is 16 29/36 or 16.81

Step-by-step explanation:

Volume of a box v = Length l × width w × height h

l = 3 1/3 = 10/3, b = 2 3/4 = 11/4, h = 1 5/6 = 11/6

v = 10/3 × 11/4 × 11/6

v = (10*11*11)/(3*4*6)

= 1210/72

V = 605/36

≈ 16 29/36 or 16.81

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A 3 foot wide brick sidewalk is laid around a rectangular swimming pool. The outside edge of the sidewalk measures 30 feet by 40
ipn [44]

Answer with Step-by-step explanation:

A 3 foot wide brick sidewalk is laid around a rectangular swimming pool.

The outside edge of the sidewalk measures 30 feet by 40 feet.

Length of swimming pool=(30-3-3) feet

                                          =24 feet

Breath of swimming pool=(40-3-3) feet

                                         = 34 feet

Perimeter of swimming pool=2(24+34) feet

(since perimeter of rectangle=2(l+b) where l is the length and b is the breath of the rectangle)

Perimeter of swimming pool=116 feet

Hence, Perimeter of swimming pool is:

116 feet

7 0
3 years ago
PLEASE HELP IN ALGEBRA
Aleksandr [31]

Sound, earthquakes, the brightness of stars, and chemistry, such as pH balance, a measure of acidity and alkalinity, are all instances of this.

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2 years ago
There are 5 red marbles, 8 blue marbles, and 12 green marbles in a bag.
lesya692 [45]

Answer:

is 28 because is probablity

Step-by-step explanation:

yes

7 0
2 years ago
Read 2 more answers
Please help quick it is due by 12 am
Anna11 [10]
Don’t you have to flip the sign when you divide by a negative? So I think she’s wrong bc the answer should be d is less than or equal to 7.5
6 0
2 years ago
Find a function where f(0)=2 and f(1)=2
xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

f(x) = 2 is a valid solution.

Want something more fun? Why not a parabola? f(x)= ax^2+bx+c.

At this point you have three parameters to play with, and from the fact that f(0)=2 we can already fix one of them, in particular c=2. At this point I would recommend picking an easy value for one of the two, let's say a= 1 (or even a=-1, it will just flip everything upside down) and find out b accordingly:f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1

Our function becomes

f(x) = x^2-x+2

Notice that it works even by switching sign in the first two terms: f(x) = -x^2+x+2

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: f(x) = A cos (kx)

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need A= 2, and at that point the first condition is guaranteed; using the second to find k we get 2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi

f(x) = 2cos(2\pi x)

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

f(x)= 2+sin(2\pi x)

Sky is the limit.

8 0
2 years ago
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