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Alborosie
3 years ago
10

Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:

Mathematics
1 answer:
valentinak56 [21]3 years ago
3 0
Here's a graph.

Part A: The graph is a shaded region with one corner and two edges that goes forever. It's only located on the right side of the y axis.

Part B: It's not a solution because it's not in the graph.

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Suppose PV =NRT, where P=8, V=2X, N=2 and R=2X.you are given X=3. Solve for T
FromTheMoon [43]

Answer:  T = 4

Step-by-step explanation:

1. Write all the variables down

P = 8  V = 2X  N = 2  R = 2X  X = 3

2. Since you know that X = 3 substitute it in to find V and R

V = 2X = 2(3) = 6

R = 2X = 2(3) = 6

3. Find PV

PV = P x V

     = 8 x 6

     = 48

4. Find NRT

NRT = N x R x T

       = 2 x 6 x T

       = 12 x T

       = 12T

5. Find T

PV = NRT

48 = 12T

12T = 48

divide both sides by 12

  T = 48 ÷ 12

  T = 4

7 0
3 years ago
Solve the equation <br> -13/4 -x= 1/2x -1
ale4655 [162]

Answer:

x = 2 or 1/4

Step-by-step explanation:

-13/4 -x= 1/2x -1

Collect like terms

-13/4+1=1/2x+x

Using LCM

(-13+4)/4=(1+2x²)/2x

9/4=(1+2x²)/2x

Cross multiply

9(2x)=4(1+2x²)

18x=4+8x²

Turn into quadratic and solve

8x²-18x+4

Using formulae method

-b±(√b²-4ac)/2a

Where a=8, b= -18 and c=4

(-(-18)±(√(-18)²-4(8)(4))/2(8)

(18±(√324-128))/16

(18±√196)/16

(18±14)/16

(18+14)/16 or (18-14)/16

32/16 or 4/16

2 or 1/4

4 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2
Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

but the integrand is even, so this is really just

\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

Substitute x = 1/2 arccot(u/2), which transforms the integral to

\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du

There are lots of ways to compute this. What I did was to consider the complex contour integral

\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}

and it follows that

\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}

7 0
2 years ago
The center of a circle is at the origin. and endpoint of a diameter of the circle is at (-3, -4. how long is the diameter of the
Aleks [24]
The diameter is 10:

To do this we must use the distance formula:

Distance =√(x2−x1)^2+(y2−y1)^2

So, if we substitute in our values for the origin and endpoint (origin is 1 values, endpoint is 2)

D=✓(-4-0)^2+(-3-0)^2

Simplified, this is

D=✓16+9
D=✓25
D=5

so, the distance from the center of the crcle to the endpoint is 5 (making the radius)

multiply by two, and the diameter of the circle is 10 :)
4 0
3 years ago
What is the scale factor from Figure 2 to Figure 1?
Luda [366]

Answer:

3

Step-by-step explanation:

Each of the lines lengths multiply by 3.

6 0
3 years ago
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