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2 years ago

What special marks are used to show that segments are congruent

Mathematics

2 answers:

ch4aika [34]2 years ago

7 0

The special marks that are used to show that segments are congruent would be: Hashmarks.

VladimirAG [237]2 years ago

4 0

Answer:

Hatch marks or hash marks are used to show that segments are congruent

Step-by-step explanation:

On any geometric figure, the congruency is denoted by hatch marks or hash marks.

For example - the notation to show the congruent sides of a triangle is done by using small tick marks, that show that the two sides are the same length or congruent.

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Four friends go to the movies and pay $11 per ticket. They also each buy a snack

Setler79 [48]

Answer:

Each snack combo costed $8

Step-by-step explanation:

4 friends* 11 per ticket = $44 for four tickets

76 - 44 = $32 spent on all the snack combos

$32 ÷ 4 friends = $8 spent on each snack combo

So, each snack combo costed $8

Hope this helps!

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2 years ago

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Solve for b. <br>4/3b+2=4-1/3b

Maksim231197 [3]

Answer:

b = $\frac{6}{5}$

Step-by-step explanation:

Multiply through by 3 to clear the fractions.

4b + 6 = 12 - b ( add b to both sides )

5b + 6 = 12 ( subtract 6 from both sides )

5b = 6 ( divide both sides by 5 )

b = $\frac{6}{5}$

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3 years ago

Find the value of the expression (5-3) x 2+6

Assoli18 [71]

Answer:

Its value is 10.

Step-by-step explanation:

Use order of operations, PEMDAS.

The parentheses are worked out first:-

(5 - 3) X 2 + 6

= 2 X 2 + 6

Now the multiplication:-

= 4 + 6

= 10 (answer).

5 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 $m_{FG}=\frac{c-0}{a+b-0}$

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

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3 years ago

WILL MARK YOU BRAINLIEST

V125BC [204]

Answer:

ASA

Step-by-step explanation:

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2 years ago

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