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brilliants [131]
3 years ago
15

On a coordinate plane, the x-coordinate is negative and the y-coordinate is positive in which quadrant?

Mathematics
1 answer:
Nataly [62]3 years ago
6 0
When x is lesser than 0
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What’s (2^2+8)x7 please help
Ivenika [448]

Answer:

84

Step-by-step explanation:

2^2=4

4+8=12

12*7=84

3 0
3 years ago
The value of y varies directly with x if x=8 then y=96 what is the value of x when y=196
ExtremeBDS [4]

Answer:

13

Step-by-step explanation:

6 0
2 years ago
Put them in the order from least to greatest. You can use the number line to help.
notka56 [123]

Answer:

-1.85 -7/8 1 3/4

Step-by-step explanation:

It is about which one is to the farest right ▶️ from 0

8 0
3 years ago
This is due NOW and I’m FREAKING OUT OMG PLEASE HELP......
Wewaii [24]

Answer:

13: Skew

14: Parallel

16: Intersecting

17: Skew

18: No

19: Yes

20: No

21: Yes

Step-by-step explanation:

Skew is when 2 lines dont intersect, nor are parallel (On different planes)

Parallel is when 2 lines are on the same plane, and will never touch

Intersecting is when any set of given points go through each other

8 0
3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term 2x^2, which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form  f(x)=ax^2+bx+c is given by the expression:

x_v=\frac{-b}{2a}

Since in our case a=2 and b=-3, we get that the x-position of the vertex is:

x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6

See the graph produced in the attached image.

4 0
3 years ago
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