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Westkost [7]
3 years ago
13

Find the volume of each figure. Round to the nearest tenth if necessary.​

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

112

Step-by-step explanation:

The volume is given by l*b*h=4*4*7=112

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write a relation that consisting of 5 ordered pairs, a domain of perfect squares and a range of perfect cubes
Aleks [24]

Answer:

Step-by-step ---------------explanation:

3 0
2 years ago
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It took 7 hours to mow four equal-size lawns. At that rate, how many lawns could be mowed in 35 hours?
NISA [10]

Answer: 20 lawns could be mowed in 35 hours.

Step-by-step explanation: Lets write a ratio to represent this problem. The ratio 7:4 states that for every 7 hours you will mow 4 lawns.

Now we have to figure out how many lawns we will mow in 35 hours. Simply keep adding 7 to the left side of the ratio and 4 to the right side.

7:4

14:8

21:12

28:16

35:20

So in 35 hours, 20 lawns will be mowed.

7 0
3 years ago
$500,3.75%,4 Months Use I=Prt
Kitty [74]
Given:
P = $500
r = 3.75% = 0.0375
t = 4 months

Asked: I
Solution: 

I = Prt
I = ($500)(0.0375)(4)

I = $75



7 0
3 years ago
Are these numbers linear or non-linear <br> x- 0, 1, 2, and 3 <br> y- 1, 3, 6, and 10
MrRissso [65]
These numbers are linear
4 0
3 years ago
Recursive definitions for subsets of binary strings.Give a recursive definition for the specified subset of the binary strings.
dmitriy555 [2]

Answer:

Step-by-step explanation:

A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your

b2n+32n+3 = 2 (2n+2)22 b2n+12n+1

your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:

1. Either placing a 1 in available n places (because you can't place it at the end)

2. or by placing a zero in available n+1 places.

0 ϵ P

x ϵ P → 1x ϵ P , x1 ϵ P

x' ϵ P,x'' ϵ P → xx'x''ϵ P

3 0
2 years ago
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