All categories

3 years ago

I need friends Drop you snap,insta,Pinterest,VSCO,email whatever of the listed things

Mathematics

2 answers:

anyanavicka [17]3 years ago

5 0

Answer:

my e m a i l is b22512312 (g m a i l . c o m) and if you email me i can tell you my insta :)

fomenos3 years ago

4 0

Snap - ashley_vicent

You might be interested in

(i will give brainliest for the first answer)

labwork [276]

Answer: X" ( -3,2) Y" (-1,0) Z" (1,6)

Step-by-step explanation:

3 0

2 years ago

Find the distance between the two points. Round to the nearest tenth, if necessary. (Point 1 is at (-5,2) and point 2 is at (3,-

aivan3 [116]

Answer:

d = 9.4

Step-by-step explanation:

d= $\sqrt{(3-(-5))^2 +(-3-2)^2\\$

d= $\sqrt{(8)^2+(-5)^2}$

d= $\sqrt{64+25}$

d= $\sqrt{89$

d=9.4

3 0

3 years ago

Solve 2x-8=-2-8. How do I do this

faust18 [17]

Is that negative 2 minus 8? if so put that into the calculator and add 8 to both sides and divide by 2 after that. simple algebra

5 0

3 years ago

Read 2 more answers

Question 10(Multiple Choice Worth 5 points)

MAVERICK [17]

So u have-27, positive 36

4 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