All categories

3 years ago

100 inches is greater than or less than or equal to 3 yard and 1ft

1 answer:

5 0

Ok so first, I am going to make an equation like this one:
100_ 3 yard and 1 ft
so I need to find out what goes into the blank so I know that there are 12 inches in 1 ft so my 1 ft is equal to 12 inches.
I also know that there are 3 ft in 1 yard so I am going to multiply 3 × 3 = 9 So know i know that there are 9 ft but now, I am going to add my 1 ft so there will be 10 ft in total. So I am now going to multiply 12 in × 10 ft To get my answer and 12 in × 10 ft = 120 inches.

So now I am going to do this:
100 inches < 120 inches ( or 3 yards and 1 ft)

You might be interested in

Rajit buys 4 pairs of socks for $2.50 per pair. He pays with a $20 bill.

sveta [45]

2.50 x 4 = 10

20 - 10 = 10

so his change would be $10

5 0

3 years ago

Read 2 more answers

Find the equation of the line that passes through the point (-2,7) and is perpendicular to the line y= 2/5x-1/5

-BARSIC- [3]

Step-by-step explanation:

ddhdhdhdhdhdddhdshshshhshshhdhhebebeewshehehehjssdjsjdudsshsusjssowossjjjddjdjddjjjs

3 0

3 years ago

It is known that a cable with a cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the

vesna_86 [32]

Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in

5 0

3 years ago

Read 2 more answers

Suppose you walk down a school hallway at a rate of 5 ft/s, and you reach your destination in 20 seconds. How do you write a fun

8090 [49]

Answer:

Step-by-step explanation:

Let's write a equation.

5 ft = 20 seconds

(five feet is reachable in 20 seconds)

Divide both sides by 5.

5/5=1

20/5=4

1 ft = 4 seconds

Let's say the destination is d and f is foot.

d = 4f

7 0

3 years ago

Read 2 more answers

Find the imaginary solutions to the following quadratic function:

Temka [501]

Answer:

the solutions are these

Step-by-step explanation:

$\frac{3}{14} + \frac{ \sqrt{159} }{14} i \: and \\ \: \frac{3}{14} - \frac{ \sqrt{159} }{14} i$

8 0

3 years ago