Please help me out. What are these ??

There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con

grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

$\angle BAC = \angle DCA$

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$ or $\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$ --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0

2 years ago

so There are 17 girls and 119 boys on a class field trip to the museum. The students need to be broken up into the largest possi

Step2247 [10]

1 student in each group with 126 groups in total

8 0

2 years ago

A baseball team's total wins and losses for one season is 162. The number of wins is 6 less than 2 times the number of losses. W

aivan3 [116]

Answer:

x + y = 162 and x = 2y - 6

Step-by-step explanation:

Let the total wins be x

Let the total loss be y

If a baseball team's total wins and losses for one season is 162, this can be expressed as;

x + y = 162 ....... 1

If the number of wins is 6 less than 2 times the number of losses, this is expressed as;

2 times the number of losses = 2y

6 less than 2 times the number of losses = 2y - 6

Now, If the number of wins is 6 less than 2 times the number of losses this is expressed as;

x = 2y - 6 .... 2

Hence the required system of equation are;

x + y = 162 and x = 2y - 6

8 0

2 years ago

URGENT!! Really need help on this test

Anuta_ua [19.1K]

Answer:

x=17

Step-by-step explanation:

53+(2x+3)=90

7 0

2 years ago

Can someone please help me solve this?

cestrela7 [59]

Answer:

(a)

Step-by-step explanation:

The tangent and the normal at point P are perpendicular.

Given

$m_{tangent}$ = 3, then

Given the gradient m of the tangent then the gradient of a line perpendicular to it is

$m_{normal}$ = - $\frac{1}{m}$ = - $\frac{1}{3}$ → (a)

3 0

2 years ago

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