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Ipatiy [6.2K]
3 years ago
15

How do I solve for y in the equation 2(3/4)+y=4?

Mathematics
2 answers:
yuradex [85]3 years ago
5 0
2* .75 + y = 4
1.5 + y = 4
Subtract 1.5 From Each Side
Y = 2.5 Or 2 1/2
beks73 [17]3 years ago
3 0
Yea the other person is right.. the answer is Y= 2.5 or 5/2 or 2 1/2
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Find the missing measures of the following rectangle
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Answer:

see explanation

Step-by-step explanation:

In a rectangle

• All angles are right angles

• the diagonals are congruent

In Δ WXV the sum of the 3 angles = 180°

∠ WXZ = ∠ XWY = \frac{180-64}{2} = \frac{116}{2} = 58°

∠ YXZ = 90° - 58° = 32°

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NEED ASAP plz. I will also mark brainiest! Don't just take points, I will report you.
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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.

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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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