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Gala2k [10]
2 years ago
10

The sum of 3 consecutive odd number is 21.What is the number.​

Mathematics
1 answer:
icang [17]2 years ago
6 0

Answer:

The numbers are 5,7,9.

Step-by-step explanation:

The numbers are 5,7,9.

5 + 7 + 9 = 21

21 = 21.

Hope this helps uwu

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A glass of water was seven- eighths of a liter. With a 6 liter jug, how many glasses could you pour? Do you have to multiply div
ivolga24 [154]

Let us suppose, we pour x glass of water.

The given jug is of 6 liters. And the glass of water was seven- eighths of a liter. Hence, we have the below equation

\frac{7}{8} x= 6

Now, in order to solve for x, we will divide 6 with 7/8. So, we have to divide.

x=\frac{6}{7/8} \\
\\
x=\frac{48}{7} \\
\\
x=6.9

Hence, we can pour 7 glass of water. But the 7th glass is not full of water. It is partially full.

5 0
3 years ago
In which number does the digit 2 have a value that is ten times as great as the value of the digit 2 in the number 135,284?
Valentin [98]

Answer: See explanation

Step-by-step explanation:

You didn't give the options but let me help out.

First and foremost, we should note that the value of the digit 2 in the number 135,284 is 200. Therefore, the value that is ten times as great as the value of the digit 2 in the number 135,284 would be:

= 10 × 200

= 2000

Therefore, the 2 must be in the thousands place. This can be seen in numbers such as:

172748.

6 0
2 years ago
Find the domain of the function y = 3 tan(2/3x)
Damm [24]

Answer:

{x∈R | \frac{2x}{3\pi } +\frac{1}{2}, x∉Z}

Step-by-step explanation:

Given the function y=3tan(2/3x)

We know that tangent is a function that's continuous within it's domain but not continuous on all real numbers

Also, the roots of y=3tan(2/3x) is 3\pi n/2 where n is an integer

Note that the domain of the function cannot be within 3\pi n/2

Therefore, {x∈R | \frac{2x}{3\pi } +\frac{1}{2}, x∉Z}

5 0
2 years ago
Particle 1 of charge q1 �� ��5.00q and particle 2 of charge q2 �� ��2.00q are fixed to an x axis. (a) as a multiple of distance
Naya [18.7K]

<span>Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0  
d = 2.72075922005613 
d = 0.612574113277207 </span>

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

3 0
3 years ago
Simplify the following expression.<br><br> ‐3 + 7(a – 3b – 1) – 4(10 – a + 2b)
mihalych1998 [28]

Answer:

11a − 29b − 50

Step-by-step explanation:

Subtract 7 from − 3

7a − 21b − 10 − 40 + 4a − 8b

Add 7a and 4a

11a - 21b - 10 - 40 - 8b

Subtract 8b from -21b

11a - 29b - 10 - 40

Subtract 40 from -10

11a - 29b - 50

7 0
2 years ago
Read 2 more answers
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