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3 years ago

What is the midpoint of the segment below?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

5 0

Answer:

Step-by-step explanation:

(5+(-4))/2 = 1/2 or 0.5

(-7 + 6)/2 = -1/2 or -0.5

the solution is D

(0.5, -0.5)

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Three cards are chosen from a standard deck of 52 playing cards with replacement what is the orobsbility every card will be a he

Ne4ueva [31]

1/4 x 1/4 x 1/4 = 1/64

There is a 1/64 chance that they will all be hearts.

7 0

4 years ago

What are the odd numbers

Ivan

Answer:

1,3,5,7,9....up to soon are odd numbers

Step-by-step explanation:

they are not divided by 2

4 0

4 years ago

Read 2 more answers

HELLPPPPPPPPPPPPP Solve x2 - 16x + 60 = -12 by completing the steps. First, subtract 60 from each side of the equation. Next, ad

kompoz [17]

Answer:

x = 8 ± 2i $\sqrt{2}$

Step-by-step explanation:

Given

x² - 16x + 60 = - 12 ( subtract 60 from both sides )

x² - 16x = - 72

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 8)x + 64 = - 72 + 64, thus

(x - 8)² = - 8 ( take the square root of both sides )

x - 8 = ± $\sqrt{-8}$ = ± 2i $\sqrt{2}$ ( add 8 to both sides )

x = 8 ± 2i $\sqrt{2}$

3 0

4 years ago

Read 2 more answers

Look at the graph below. Which of the following best represents the slope of the line? A. -3 B. - 1 3 C. 1 3 D. 3

Salsk061 [2.6K]

Answer and Explanation:

The line seems to pass trough the points (0,6) and (-3,4).

Slope is: $m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{4-6}{-3-0} =\frac{-2}{-3} =\boxed{\frac{2}{3}}$

The slope of the line should be $\frac{2}{3}$ .

3 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