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Paha777 [63]
3 years ago
12

Solve 4(t+1)=6t-1 A) 1 1/2 B) 2 1/2 C) 1 D) 0

Mathematics
2 answers:
Llana [10]3 years ago
4 0

4(t+1)=6t-1

4t+4=6t -1

4t-6t+4=6t-6t-1

-2t+4=-1

-2t+4-4=-1-4

-2t=-5

Divide by -2 for -2t and -5

-2t/-2=-5/-2

t=5/2= 2 1/2

Answer is b) 2 1/2

-BARSIC- [3]3 years ago
3 0

Answer:

answer is B...2 1/2

Step-by-step explanation:

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Answer:

≈ 0.2526

Step-by-step explanation:

<u>The number of combinations of 4 out of 8:</u>

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Step-by-step explanation:

1) A number <em>m</em> is at least four

<em>m ≥ 4  ['at least' = ≥ symbol]</em>

<em />

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[Im not sure if this is a solution or and inequality :/ ]

3)  x < 5

--> On the line, draw the number 0, and the number five. 'x < 5' also means 'x is less that five.' So, draw a dot over the line before 5, and make an arrow going to the left of the page,

<-----------------------------------o   5

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Part A:

You may choose the two lines connecting the origin and points A and B, and choose the portion of the space between them.

The line between the origin and A is

y = 3x

We want everything below this line (line included), so the first inequality is

y \leq 3x

The line between the origin and B is

y = \dfrac{1}{3}x

We want everything above this line (line included), so the second inequality is

y \geq \dfrac{1}{3}x

Create a system with these two inequalities and you'll have an area including only points A and B

Part B:

To verify the solutions, we can plug the coordinates of A and B in this system and check that we get something true: the coordinates of point A are (1,3), while the coordinates of point B are (3,1). The system becomes:

A:\begin{cases}3 \leq 3\cdot 1\\3 \geq \frac{1}{3}\cdot 1\end{cases},\quad B:\begin{cases}1 \leq 3\cdot 3\\1 \geq \frac{1}{3}\cdot 3\end{cases}

Which means

A:\begin{cases}3 \leq 3\\3 \geq \frac{1}{3}\end{cases},\quad B:\begin{cases}1 \leq 9\\1 \geq 1\end{cases}

And these are all true. So, the system is satisfied, which means that the points belong to the shaded area.

Part C

If you draw the line, you'll see that the only points that lay below the line are B and C. In fact, if we plug the coordinates we have

B:\ 1

And this are both true. You can check the coordinates of all other points, and see that they won't satisfy the inequality y<3x-6

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Answer:

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