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valentinak56 [21]

3 years ago

11

Choose the system of inequalities that best matches the graph below.

1 answer:

lapo4ka [179]3 years ago

4 0

Answer:

D) $y\:$ , $y\:>\:-x$

Step-by-step explanation:

The blue boundary line has a y-intercept of 2 and a slope of 2.

It has equation : $y=2x+2$

Since the boundary line is not solid and the lower half plane of this line is shaded, its inequality is

$y\:$

The blue boundary line passes through the origin and has slope $-1$ .

Its equation is $y=-x.$

Since the boundary line is not solid and the right half-plane is shaded, its inequality is $y\:>\:-x$

The correct choice is D.

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5 0

2 years ago

What is the greatest value 20g of flour 3 for 3.50 or 40g of flour for 2 for 9£ which is better for money

jenyasd209 [6]

Answer:

It is better to buy 20 g of flour 3 for 3.50

Step-by-step explanation:

It is given that 20g of 3 for 3.50

It means 3 packets of 20 g is 3*20 =60

So, 60g is for 3.50

So, 1 g = $\frac{3.5}{60}$ =0.05833 per gram.

Now, let's find unit rate for 40 g of 2 for 9.

2 packets of 40 gram = 2* 40 =80 grams

So, 80 grams costs 9

So, 1 gram = $\frac{9}{80}$ =0.1125

So, it is better to buy 20 g of flour 3 for 3.50

4 0

2 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

2 years ago

Find f^-1(x) for f(x)=1/x^3

Nata [24]

if you love god give me 5 stars.................

4 0

2 years ago

What is the most precise name for quadrilateral ABCD with vertices A(-4, -4), B(-4, -2), C(-1, -2), and D(-1, -4)?

baherus [9]

D a rectangle because it has four sides easy for abcd

5 0

3 years ago

Read 2 more answers