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blsea [12.9K]
3 years ago
13

Complete the proofs using the most appropriate method. may require CPCTC.

Mathematics
1 answer:
n200080 [17]3 years ago
3 0

For the first one, you did good. I will just suggest a couple things.

Statement              Reason

JK ≅ LM                   Given

<JKM ≅ < LMK        Given  (You did both of these steps so well done.)

MK ≅ MK              Reflexive Property (Your angle pair is congruent but isn't one of the interior angle of the triangles you are trying to prove.)

ΔJMK ≅ ΔLKM        SAS

Problem 2: (You also have a lot of great stuff here.)

Statement          Reason

DE ║ FG              Given

DE ≅ FG              Given

<DEF≅<FGH       Given

<EDF≅<GFH       Corresponding Angles (You don't need to know that F is the midpoint but you got corresponding angle pair which is correct.)

ΔEDF≅ΔGFH        ASA

<DFE≅<FHG       CPCTC  

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4 0
3 years ago
Please help me understand this
Keith_Richards [23]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Jerry’s rhombus has a base of 15 inches and a height of 9 inches. Charlene’s rhombus has a base and a height that are 13 the bas
neonofarm [45]

Answer:  The area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

Step-by-step explanation:

I will assume that the exercise says "\frac{1}{3} times the base and height of Jerry’s rhombus".

The area of a rhombus can be calculated with the following formula:

A=b*a

Where "b" is the length of the base  and "a" is the altitude or the height.

Then, you can calculate the area using the formula shown above.

Therefore, you get:

1. Jerry's rhombus:

A_1=(15\ in)(9\ in)\\\\A_1=135\ in^2

2. Charlene's rhombus:

A_2=(\frac{1}{3}*15\ in)(\frac{1}{3}*9\ in)\\\\A_2=(5\ in)(3\ in)\\\\A_2=15\ in^2

Dividing the area calculated, you get:

\frac{135\ in^2}{15\ in^2}=9

Therefore, you can conclude that the area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

7 0
4 years ago
Read 2 more answers
PLEASE HELP/ANSWER! From 1980 to 1990, Lior’s weight increased by 25%. If his weight was k kilograms in 1990, what was it in 198
Elena-2011 [213]

The weight in 1980 is \frac{4k}{5} kilograms

<em><u>Solution:</u></em>

From 1980 to 1990, Lior’s weight increased by 25%

His weight is "k" kilograms in 1990

<em><u>To find: weight in 1980</u></em>

This is a percentage increase problem

Let "x" be the weight in kilograms in 1980

<em><u>The percentage increase is given by formula:</u></em>

\text{Percentage increase } = \frac{\text{Final value - initial value}}{\text{initial value}} \times 100

Here,

Initial value in 1980 = x

Final value in 1990 = k

Percentage increase = 25 %

<em><u>Substituting the values in formula,</u></em>

25 = \frac{k-x}{x} \times 100\\\\25x = 100(k-x)\\\\x = 4(k-x)\\\\x = 4k - 4x\\\\5x = 4k\\\\x = \frac{4k}{5}

Thus the weight in kilograms in 1980 is \frac{4k}{5}

7 0
4 years ago
Simplify to create an equivalent expression<br><br> 6(5r-11)-(5-r)
dolphi86 [110]

Answer:

B or 31r-71

Step-by-step explanation:

First, use distributive property to expand the first term.

30r-66-(5-r)

Remove the parenthesis of the third term:

30r-66-5+r

Combine like terms

31r-71

7 0
4 years ago
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