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WINSTONCH [101]

3 years ago

6

PLEASE HELP ME WITH THIS QUESTION, ILL BRAINLIEST YOU

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

i think 105 is the correct answer Hope it helps you have a great day keep smiling be happy stay

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Find the x intercept for the graph, 3x-14y+12=0

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The x-intercept would be (-4,0)

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3 years ago

What is another way to write the expression p • (10 - 2)

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Answer:

(p • 10) - (p • 2)

Step-by-step explanation:

To be doubly sure of your answer, do the actual mult.:

p(10-2) = 10p - 2p. This is equivalent to (p • 10) - (p • 2) (Answer D).

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What is the solution to x ? <br><br> 7/3x + 1/3x = 3x + 6/3 + 5/3x

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3 years ago

What is the inverse of y=x+5

V125BC [204]

Answer:

x=-5

Step-by-step explanation:

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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