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3 years ago

Why did the Vietnam vets feel alienated when they came home from war

Mathematics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

Because they were suffering form PTSD and questioning their identity.

You might be interested in

2/0 = 32/48 = 30/0. What number shpuld come in place of the zeros?

Anton [14]

Answer:

$\sf \dfrac{2}{3} = \dfrac{32}{48} = \dfrac{30}{45}$

$\sf \rightarrow \dfrac{2}{x} = \dfrac{32}{48}$

$\sf \rightarrow 2(48)= {32x}$

$\sf \rightarrow 32x= 96$

$\sf \rightarrow x = 3$

$\rightarrow \sf \dfrac{32}{48} = \dfrac{30}{y}$

$\rightarrow \sf {32y} = 30(48)$

$\rightarrow \sf {32y} = 1440$

$\rightarrow \sf y = 45$

8 0

2 years ago

At a dinner at Priya's favorite restaurant, the meal cost $18 and a sales tax of $1.71 was added to the bill.

Naily [24]

Answer:

$3.23

Step-by-step explanation:

1.71 ÷ 18 = 0.095 = 9.5%

9.5% × 34 = 3.23

5 0

2 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

3 years ago

Which region labeled in the graph below would represent the solution (the final shaded area) to the system of linear inequalitie

Elena L [17]

Answer:

Region D.

Step-by-step explanation:

Here we have two inequalities:

y ≤ 1/2x − 3

y < −2/3x + 1

First, we can see that the first inequality has a positive slope and the symbol (≤) so the values of the line itself are solutions, this line is the solid line in the graph.

And we have that:

y ≤ 1/2x − 3

y must be smaller or equal than the solid line, so here we look at the regions below the solid line, which are region D and region C.

Now let's look at the other one:

y < −2/3x + 1

y = (-2/3)*x + 1

is the dashed line in the graph.

And we have:

y < −2/3x + 1

So y is smaller than the values of the line, so we need to look at the region that is below de dashed line.

The regions below the dashed line are region A and region D.

The solution for the system:

y ≤ 1/2x − 3

y < −2/3x + 1

Is the region that is a solution for both inequalities, we can see that the only region that is a solution for both of them is region D.

Then the correct option is region D.

5 0

3 years ago

How??????????????????????

Digiron [165]

Answer:

y=-1/3x+7

Step-by-step explanation:

y=mx+c

m=-1/3, c=7

y=-1/3x+7

5 0

3 years ago