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3 years ago

5

Please help me im having so much trouble

1 answer:

Karolina [17]3 years ago

5 0

It would be B, an acute scalene. It would be B because all 3 sides are different.

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The scatter plot shows the relationship between the test scores of a group of students and the number of hours they spend on soc

Rasek [7]

Hi there!

Part A: Those groups of points that are circled and labeled A are called a correlation. In this case, the group of points represent negative correlation. The point that is labelled B is called an outlier, because it falls outside of the range of the other points. Point B may have been present like that, because while a person may spend a lot of time on social media, they might have understood the information very fast, or maybe they studied a bit on a different occasion.

Part B: Going back to part A, the graph represents a negative correlation. In this case, the more time spent on social media, the lower the test scores.

7 0

3 years ago

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

3 years ago

Find the midpoint of the segment with the endpoints:(-7,-8) and (4,7)

Oxana [17]

Answer: (-1.5, -0.5)

Step-by-step explanation:

8 0

3 years ago

Which equation is solved for the variable c in the equation a(b-c)=d

Nady [450]

C= (-d/a) +b Hope this helps

7 0

3 years ago

The number of exports is shown as g(x). Use the data in the table below, representing both functions, to explain to your boss th

siniylev [52]

Step-by-step explanation:

Given the table:

Month f(x) = Number of imports g(x) = Number of exports

January (1) 3 1

February (2) 4 3

March (3) 5 5

April (4) 6 7

From the table, it is clear that:

Number of imports = number of month + 2

so

$f(x) = x + 2$

also

Number of exports = 2(number of month) + 1

so

$g(x) = 2x - 1$

Therefore, number of imports equals the number of month plus one. i.e. $f(x) = x + 2$ , which is a linear function.

If we compare it with slope-intercept form of the line

$y=mx+b$

Then,

slope = m = 1

y-intercept = 2

Also number of exports equals is also a linear function.

i.e. $g(x) = 2x - 1$

Here,

slope = m = 2

y-intercept = -1

3 0

2 years ago