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xenn [34]
3 years ago
5

On a piece of paper , graph y>2x-3. Then determine which answer choice matches the graph you drew

Mathematics
2 answers:
irakobra [83]3 years ago
8 0

Answer:

Graph represented by option C is the correct choice.

Step-by-step explanation:

We are asked to graph an inequality y\geq2x-3.

The boundary line of our given inequality will be a solid line as we have greater than or equal to \geq sign.

The boundary line of our given inequality would be y=2x-3.

Now, we will test point (0,0) to shade in the correct region as:

0\geq2(0)-3

0\geq0-3

0\geq-3

Since the point (0,0) is a solution for our given inequality, therefore, the shaded area of our inequality will include point (0,0).

Please find the attachment.

Damm [24]3 years ago
6 0

Answer:

C

Step-by-step explanation:

Since it is greater than or equal to, the graph line is a solid line.

Then, since we know that it is greater than, we know that we can shade above the graph line.

To check this, all you have to do is choose a random point in the shaded region and plug it in the equation. If it checks out, it is correct.

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PLEASE HELP ASAP <br><br> Thank you!!! &lt;3
Crazy boy [7]
2.625e14. now all u have ro do is simplify it
6 0
3 years ago
Helpppppppppppppppppppp!
Advocard [28]

It looks like this is a square pyramid net, and to find the surface area you need to know the base and height. By looking at the image 5 ft is the height and the base is 3 ft.

The surface area of the square pyramid is <u>40.32</u>

hope this helps!

5 0
3 years ago
Using 4 equal-width intervals, show that the trapezoidal rule is the average of the upper and lower sum estimates for the integr
prisoha [69]

Split up the interval [0, 2] into 4 subintervals, so that

[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]

Each subinterval has width \dfrac{2-0}4=\dfrac12. The area of the trapezoid constructed on each subinterval is \dfrac{f(x_i)+f(x_{i+1})}4, i.e. the average of the values of x^2 at both endpoints of the subinterval times 1/2 over each subinterval [x_i,x_{i+1}].

So,

\displaystyle\int_0^2x^2\,\mathrm dx\approx\dfrac{0^2+\left(\frac12\right)^2}4+\dfrac{\left(\frac12\right)^2+1^2}4+\dfrac{1^2+\left(\frac32\right)^2}4+\dfrac{\left(\frac32\right)^2+2^2}4

=\displaystyle\sum_{i=1}^4\frac{\left(\frac{i-1}2\right)^2+\left(\frac i2\right)^2}4=\frac{11}4

4 0
3 years ago
Please help i’m finding this hard
pav-90 [236]

Answer:

Area for Shape p is 12, the name for the shape q is a rhombus.

Step-by-step explanation:

The shape p is a triangle for if you find the height of the shape which is 6 and the length of the base which is 4 multiply them together then divide by 2 to get the area.

8 0
3 years ago
What is the solution set for
lana [24]
It’s s=-2 and s=2 because the absolute value makes both of the numbers be positive so either way it’s going to be 2+4=6
5 0
2 years ago
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