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Gnom [1K]
3 years ago
5

Find the inequality represented by the graph. Thanks! (plz no spam...)

Mathematics
2 answers:
nikitadnepr [17]3 years ago
6 0

Answer:

y > (1/3)x - 3.

Step-by-step explanation:

The graph of a line can be written in y = mx + b form where b is the y-intercept and m is the slope.

In the image, the slope of the graph from the given points is 1/3.

In the image, the y-intercept is -3.

Therefore, the line is y = (1/3)x -3. However, we aren’t done yet! This is an inequality, not an equation!

We see the line isn’t dotted, so that means it must be > or <.

We substitute the point (0,0) into the line equation we got and find that 0 > (1/3)(0) - 3 = -3. Since (0,0) is part of the inequality, we have that y > (1/3)x - 3.

I hope this helps! :)

Leona [35]3 years ago
4 0

Answer:

y > (1/3)x - 3.

Step-by-step explanation:

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Solve using the standard algorithm.
Vinvika [58]

Answer:

A= 1.12

B= 1.11

Step-by-step explanation:

A.     0.82                 B.      1.03    

+       0.3                        +  0.08

   ______                     _______

      1.12                            1.11

8 0
3 years ago
A circle has radius 50 cm. Which of these is closest to its area? Group of answer choices
kobusy [5.1K]

Answer:

7850 cm

Step-by-step explanation:

The formula for area of a circle is: \pir^{2}

50^{2} = 2500

(\pi will be rounded to 3.14)

2500\pi = 7850

8 0
3 years ago
HELP PLEASE I NEED HELPPPP!!!!!!!!!!
Kisachek [45]

two triangles is equal to 12 ft squared

the area of rectangles is equal to 8 ft x 6 ft which will give you 48 feet squared

you add them together to get the total area and you will get 84 feet sqaured

3 0
3 years ago
The graph shows the saving in Andre's bank account. What is the slope of the line?
VLD [36.1K]

Answer:

5

Step-by-step explanation:

slope= rise/run

=y2-y1/x2-x1

Take any two points

80-40/8-0

40/8

=5

8 0
3 years ago
Consider the differential equation: (y^2+2*x)dx+(2*x*y-1)dy = 0. (a) show that the equation is exact by evaluating (\partial m)/
ruslelena [56]

The given ODE

\underbrace{(y^2+2x)}_M\,\mathrm dx+\underbrace{(2xy-1)}_N\,\mathrm dy=0

is exact if \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. It so happens that we have \frac{\partial M}{\partial y}=2y=\frac{\partial N}{\partial x}, so it is indeed exact. For such an ODE, we're looking for a solution of the form \Psi(x,y)=C, for which the differential is

\mathrm d\Psi=\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0

so we have the following system of PDEs that allow us to solve for \Psi:

\dfrac{\partial\Psi}{\partial x}=M=y^2+2x

\dfrac{\partial\Psi}{\partial y}=N=2xy-1

In the first PDE, we can integrate both sides with respect to x and recover \Psi:

\displaystyle\int\frac{\partial\Psi}{\partial x}\,\mathrm dx=\int(y^2+2x)\,\mathrm dx

\implies\Psi(x,y)=xy^2+x^2+f(y)

Then differentiating this with respect to y returns N:

\dfrac{\partial\Psi}{\partial y}=2xy+\dfrac{\mathrm df}{\mathrm dy}=2xy-1

\implies\dfrac{\mathrm df}{\mathrm dy}=-1

\implies f(y)=-y+C

So the general solution to the ODE is

\Psi(x,y)=xy^2+x^2-y+C=C

or simply

xy^2+x^2-y=C

4 0
3 years ago
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