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Neporo4naja [7]
3 years ago
11

Which inequality will have a solid boundary line and a shaded region above its graph? A. x − y ≥ 3 B. 2x − 3y ≤ 3 C. 3y − x <

2 D. 2x + y < 7
Mathematics
2 answers:
N76 [4]3 years ago
6 0

The answer is B. 2x - 3y less than or equal to 3

professor190 [17]3 years ago
4 0

Answer:

option B, 2x - 3y ≤ 3

Step-by-step explanation:

To solve the question of inequality we should remember few things.

(1)  Solid line means sign of equality

(2)  Shaded region left side on above the line means " less than"

(3) shaded region right side or below the line means "greater than"

Now from the given option B, 2x - 3y ≤ 3 has the property of "less than equal to" showing a solid line and shaded part above  its graph.

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Jack had 3 bottles of soda. Each bottle held 2 1/2 cups of soda. How many cups of soda did he have total?
lukranit [14]

Answer:

1 1/5 cups of soda.

Step-by-step explanation:

To solve this you need to change 3 and 2 1/2 into improper fractions so you get 6/2 and 5/2. Then you have to flip 5/2 so it becomes 2/5 and then you just have to cross cancel. Since the 2 from 2/5 can go into the 2 in 6/2 one time, 2/5 changes into 1/5 and 6/2 changes to 6/1. Now you can multiply across so 6 times 1 is 6 and 1 times 5 is 5 so you get 6/5. Now you have to change it into a mixed number which would be 1 1/5.

I hope this helps you :D

7 0
3 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
There are 45 children in a class. If
Dovator [93]

Answer:

20%

Step-by-step explanation:

Total = 45

Absentees = 9

Percentage  = (No. of absentees / total) × 100

                    = 9 / 45 × 100

                    = 0.2 × 100

                    = <u>20%</u>

3 0
2 years ago
Read 2 more answers
What is the volume of the triangular prism​
mestny [16]

Answer:

  • 308 cm³

Step-by-step explanation:

<u>The volume is:</u>

  • V = Bl = 1/2(ah)l
  • V = 1/2(7*8)*11 = 308 cm³
3 0
3 years ago
Read 2 more answers
in the figure below, lines a and b are straight and parallel to each other, which angles must be congruent to&lt;1
vlabodo [156]

You have to show a figure

8 0
3 years ago
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