Ask question

Ask question

All categories

3 years ago

8

Can you use the SAS Postulate, the AAS Theorem, or both to prove the triangles congruent? A. either SAS or AAS B. SAS only C. AA

S only D. neithe

1 answer:

Harman [31]3 years ago

5 0

Sas only because the triangles and usually congruent to me or in my opponion

You might be interested in

Find the GCF of 5y5 and 50y2.

GREYUIT [131]

Answer: $5y^2$

Step-by-step explanation:

Given

$5y^5\ \text{and}\ 50y^2$

GCF or HCF of the two is the highest common factor

$\Rightarrow 5y^5=5\times y^2\times y^3\\\Rightarrow 50y^2=5\times y^2\times 10$

It is clear GCF is $5y^2$

6 0

3 years ago

Which of the following sequences of numbers are arithmetic sequences?

serg [7]

Hello,

A:no
B:no
C,D,E :yes

All you have to do : is the difference of next term and the term must be constant.

5 0

3 years ago

Read 2 more answers

In an average of 65 minutes if a train stops at three stations for two minutes each and the distance is 115 miles between stops

larisa86 [58]

Answer:

117mph

Step-by-step explanation:

Given that

Total number of stations= 3

Total average time of journey of train = 65

total stop time at three stations = 3×2 =6 min

Total distance between stops = 115 miles

as we can see

Total time train was travelling = 65- 6 =59 min

so, average speed of train can be calculated as:

Average Speed = $\frac{Distance -travelled-between -stations}{Total-time-the-train-travelled}$

Average Speed = $\frac{115}{59} =1.95 miles/minute$

To convert this speed into miles per hour (mph) multiply the above speed by 60

Average speed = 60×1.95 =117 mph

So the average speed of train is 117mph.

3 0

3 years ago

Add these mixed numbers -Thanks!

alina1380 [7]

1. 6 1/2
2.4 1/2
3. 9 1/3
4. 9 1/2
5. 9 6/8
6. 6 1/2
(For the adding ones ) I also simplified them.

7 0

3 years ago

Can someone help me please

VashaNatasha [74]

Answer:

TQ = 10 $\sqrt{2}$

Step-by-step explanation:

The diagonal divides the square into 2 right triangles with the diagonal being the hypotenuse.

Using Pythagoras' identity in triangle TPQ

TQ² = PQ² + PT²

TQ² = 10² + 10² = 100 + 100 = 200 ( take square root of both sides )

TQ = $\sqrt{200}$ = $\sqrt{100(2)}$ = $\sqrt{100}$ × $\sqrt{2}$ = 10 $\sqrt{2}$

7 0

3 years ago