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PIT_PIT [208]
3 years ago
12

Solve the answer 6p - 1 - 4p = 15?

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0

Answer:

it would equal <em>2p-1 </em>as a mathematical equation

Step-by-step explanation:

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Volume equals length times width times height v=lwh<br> solve the equation for l
marshall27 [118]

Answer:

\frac{V}{wh}=l

Step-by-step explanation:

Equation: V=lwh

1). \frac{V}{wh}=\frac{lwh}{wh}

2). \frac{V}{wh}=l

3 0
3 years ago
The US postal service delivered 7.14 x 1010 pieces of mail in the month of December and 3.21 x 1010 in the month of January. Wha
kupik [55]
The US postal service delivered 7.14 x 1010 pieces of mail in the month of December 7.14 x 1010=7211.4

3.21 x 1010 in the month of January. 3242.1

What is the total mail delivered for these two months?

7211.4
+ 3242.1
__________

10453 .5 TOTAL MAIL
4 0
3 years ago
4a + 3b - 2a - 6b<br> .. help lol
ycow [4]

Answer:

2a-3b

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
Give an example of an open equation​
rusak2 [61]

the example of an open equation is :x+9=16

4 0
3 years ago
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