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andrew-mc [135]
3 years ago
5

Help please,it may be easy for you,but it's hard for me.Question may be confusing but

Mathematics
1 answer:
olga2289 [7]3 years ago
8 0
I got 37....because 180 - 143 = 37.....
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Twenty two quarts is equivalent to how many gallons?<br> 4.5<br> 5.5<br> C<br> 6.5<br> 7.5
marysya [2.9K]

Answer:

5.5 gallons

Step-by-step explanation:

There are 4 quarts in 1 gallon.

22 ÷ 4 = 5.5

I hope this helps :)

4 0
3 years ago
I need help with this pls use 3.14 with or the formula with area
Leni [432]
Rectangle:
A=l(w)

A=12(6)

A=72cm^2

Semicircle:
r=6/2

r=3

A=(3.14(r)^2(1/2)

A=3.14(3)^2(1/2)

A=3.14(9)(1/2)

A=14.13cm^2

A=14.1cm^2

Total area: 72cm^2 + 14.1cm^2 = 86.1cm^2
6 0
2 years ago
I have k quarters, five less quarters than nickels and one more than twice as many dimes as quarters. Find the value of the coin
Alenkinab [10]

Answer:

(35k + 20) cents

Step-by-step explanation:

First of all, let us have the value of each unit:

1 quarter = 25 cents

1 nickel = 5 cents

1 dime = 10 cents

Given that number of quarter = k

Quarters are 5 lesser than Nickels, so number of nickels = k+5

One more than twice as many dimes as quarters:

k = 2 \times Number of Dimes + 1

So, number of dimes = \frac{1}{2}(k-1)

Value of quarters = 25 \times k cents

Value of nickels = 5 \times (k+5) = (5k+25)\ cents

Value of dimes = \frac{1}{2}(k-1) \times 10 = (5k-5)\ cents

So, total value of coins =

25k + 5k +25 +5k-5\\\Rightarrow (35k+20)\ cents

5 0
3 years ago
Read 2 more answers
Which equation can be used to find the surface area of the cylinder?
Elan Coil [88]

Answer:

D

Step-by-step explanation:

correct

5 0
2 years ago
Read 2 more answers
The sum of three numbers is 11 . The sum of twice the first​ number, 5 times the second​ number, and 6 times the third number is
Assoli18 [71]

Answer:

4, 6, 1

Step-by-step explanation:

We can solve this problem using a system of equations:

1) a + b + c = 11

2) 2a + 5b + 6c = 44

3) 4a - b = 10

First, we can substitute the value of b from equation #3 into equation #1:

b = 4a - 10

a + (4a - 10) + c = 11

5a - 10 + c = 11

5a + c = 21

c = 21 - 5a

Now, we can plug the values of b and c into equation #2, as b and c are represented in terms of a:

2a + 5(4a - 10) + 6(21 - 5a) = 44

2a + 20a - 50 + 126 - 30a = 44

-8a + 76 = 44

-8a = -32

a = 4

b = 4a - 10 = 4(4) - 10 = 6

c = 21 - 5a = 21 - 5(4) = 1

4 0
2 years ago
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