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STatiana [176]
3 years ago
8

5 common multibles of 7 and 20

Mathematics
1 answer:
uranmaximum [27]3 years ago
3 0
7 = 7
20 = 2 x 2 x 5

Least common multiple is 2 x 2 x 5 x 7 = 140
Other multiples are multiples of 140 = 280, 420, 560, 700
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perter's glasses each hold 8 fluid ounces.how many glasses of juice can perter pour from a bottle that holds 5 quarts
NeX [460]

Answer:

8 glasses :)

Step-by-step explanation:

There is 32oz per quart. 8 goes in to 32 a total of four times. So since there are two quarts, Peter can pour 8 glasses.

From everything I've seen it should be 2 quarts from my knowledge. Sorry if this doesn't help!

3 0
3 years ago
List all the 3-digit numbers that fit these clues .the hundreds digit is greater than7 .the tens is 1 more than the hundreds dig
djverab [1.8K]
In this question, there are 3 conditions that need to be met
1. The number is 3-digit
2. Hundreds digit >7 (number that will fulfill it would be 8 or 9)
3. The tens is 1 more than hundred( since the hundred possibilities is 8 or 9, then 9 in hundreds can't be used since no number higher than 9)
4. The one digit <2( that mean 0, 1)

Using the list above, you can make your possible number:
890
891
3 0
3 years ago
Evaluate the expression 2(4-3) in two ways pls help
Morgarella [4.7K]

Answer:

Step-by-step explanation:

Method 1:

First do the operation inside parenthesis

2*(4-3) = 2*1 = 2

Method 2:

Use distributive property: a*(b +c) = a*b + a*c

2*(4-3) = 2*4 - 2*3

           = 8 - 6

            = 2

6 0
2 years ago
Which number is not in scientific notation?
Hitman42 [59]

Answer:

A 9 ⋅ 1019

B 7.8 ⋅ 106

C 1.45 ⋅ 10−7

D 0.33 ⋅ 10−15

Step-by-step explanation:

A 9 ⋅ 1019

B 7.8 ⋅ 106

C 1.45 ⋅ 10−7

D 0.33 ⋅ 10−15

6 0
3 years ago
Ana escribe conjuntos de cinco enteros positivos consecutivos con la siguiente propiedad: La suma de tres de los números es tan
Keith_Richards [23]

Answer:

a) 1.

Step-by-step explanation:

Supongamos que el primer número entero positivo es a tal que a ∈ Z⁺.

Luego, el conjunto de los cinco enteros positivos se puede expresar como

A = {a; a+1; a+2; a+3; a+4}

Dada la condición del problema, se debe cumplir que

a + (a+1) + (a+2) = (a+3) + (a+4)

⇒  3a + 3 = 2a + 7

Resolviendo la ecuación resulta

a = 4

Luego, el conjunto A nos resulta

A = {4; 5; 6; 7; 8}

Puede concluirse que sólo un conjunto cumple con esta condición.

5 0
2 years ago
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