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Y_Kistochka [10]
3 years ago
11

What is the greatest common factor of 5xy and 20x

Mathematics
1 answer:
kvasek [131]3 years ago
7 0

Answer:

5x

Step-by-step explanation:

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El perímetro de un rectángulo es de 54 pulgadas. Su longitud dos veces es ancho. Encontrar la longitud y anchura del hallazgo re
elixir [45]
Longitud = L  Ancho = A
2A + 2L = 54
L = 2A    Sustituir este valor por L en la primera ecuación:
2A + 2(2A) = 54
2A + 4A = 54
6A = 54
A = 9
También porque L = 2A,
L = 2(9) = 18
3 0
3 years ago
Use partial quotients to solve 504 divide 14
miv72 [106K]

Answer:

Perform the following division using partial quotients: 504 ÷ 14:

To reduce the numerator, we will be multiplying the denominator by factors of 10, 5, 2, and 1

Step-by-step explanation:

1 4| 5 0 4

2 8 0 20 <---- 20 x 14 = 280

2 2 4 <---- 504 - 280 = 224

1 4| 5 0 4

2 8 0 20

2 2 4

1 4 0 10 <---- 10 x 14 = 140

8 4 <---- 224 - 140 = 84

1 4| 5 0 4

2 8 0 20

2 2 4

1 4 0 10

8 4

7 0 5 <---- 5 x 14 = 70

1 4 <---- 84 - 70 = 14

1 4| 5 0 4

2 8 0 20

2 2 4

1 4 0 10

8 4

7 0 5

1 4

1 4 1 <---- 1 x 14 = 14

0 <---- 14 - 14 = 0

Our partial quotients add up as follows:

20 + 10 + 5 + 1 = 36

I hope it helps plz Mark me brainliest:)

5 0
3 years ago
Which statement best reflects the solution(s) of the equation? 1/x+1/x−3=x−2/x−3
Virty [35]
It is a i hope that help
5 0
3 years ago
Read 2 more answers
4. Diego is thinking of two positive numbers. He says, "If we triple the first number and
aleksandr82 [10.1K]

Answer:

3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are  (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).

Step-by-step explanation:

 Let the First positive number be <em>x</em> and second positive number be <em>y.</em>

Triple of first number = 3<em>x</em>

double of second number = 2<em>y</em>

According to question,

3<em>x </em>+ 2<em>y</em> = 34

Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34

So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34

Now,  3<em>x </em>+ 2<em>y</em> = 34

3<em>x </em>= 34 - 2<em>y</em>

Let x = 10 and by substituting its value in above expression,

3 \times 10 = 34 - 2y

30 = 34 - 2y

2y = 34 - 30

2y = 4

y = 2

Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)

In the same way put x = 4 then,

3<em>x </em>= 34 - 2<em>y</em>

3 \times 4 = 34 - 2y

2y = 34 - 12

2y = 22

y = 11

Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)

Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and  (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.  

8 0
3 years ago
2(x + 2) + 3z = 2(x + 1) +1
Oxana [17]

Answer:

Answer:x=

−3z−1

0

Step-by-step explanation:x=

Step 1: Add -2x to both sides.

2x+3z+4+−2x=2x+3+−2x

3z+4=3

Step 2: Add -3z to both sides.

3z+4+−3z=3+−3z

4=−3z+3

Step 3: Add -4 to both sides.

4+−4=−3z+3+−4

0=−3z−1

Step 4: Divide both sides by 0.

0

0

=

−3z−1

0

x=

−3z−1

0

6 0
3 years ago
Read 2 more answers
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