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svetlana [45]
3 years ago
12

What is 2 2/3 - 1 1/3? answer using backslash for example 2/3

Mathematics
1 answer:
Rudik [331]3 years ago
6 0

Answer:

-1

Step-by-step explanation:

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Select the system of linear equations that has the solution (-6, 0).
xxTIMURxx [149]

Answer:

B.

• x+2y=-6

• x-2y=-6

Step-by-step explanation:

Try the solution in the equations and choose the set of equations that is true.

___

A. -6+0 ≠ 0 . . . not this system

__

B. -6 +2·0 = -6 . . . true

-6 -2·0 = -6 . . . true . . . . . this system has solution (-6, 0)

__

C. -6 -6·0 ≠ 1 . . . not this system

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D. 4(-6)+3·0 ≠ -18 . . . not this system

3 0
3 years ago
Will give brainliest if done properly
k0ka [10]

Answer:

\sf  \boxed{  \sf y - 3 = \dfrac{3}{2} (x - 6)}

Explanation:

\sf Given \ equation: y = \dfrac{3}{2} x - 5

Comparing it with slope intercept form "y = mx + b" where 'm' is slope and 'b' is y-intercept.

Here slope: 3/2 and y-intercept: -5

Parallel slope has the same tangent slope.

Pass through point (x, y) = (6, 3)

Equation:

\sf y - y_1 = m(x  - x_1)

\rightarrow \sf y - 3 = \dfrac{3}{2} (x - 6)

6 0
2 years ago
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Natali5045456 [20]

Answer:

The selected answer seem correct...

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Step-by-step explanation:

3 0
3 years ago
What % of 300 is 30​
algol13

Answer:10%

Step-by-step explanation:

5 0
4 years ago
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
4 years ago
Read 2 more answers
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