Hi!
A <u>hypotenuse</u> is the longest leg in a right triangle.
A <u>right triangle</u> is a triangle that has one 90 degree angle.
The hypotenuse is always directly across from the right angle.
In pictures, right angles are represented by a square.
<u><em>Therefore, the side with the hypotenuse is side 3</em></u>, because we can see that the angle between side 1 and side 2 is the right angle, and the hypotenuse would be directly across.
<u><em>For more information, see:</em></u>
brainly.com/question/21440624
<u><em>For information about parts of a right triangle, see:</em></u>
brainly.com/question/22137749
Image = other parts of a right triangle:
Answer:
Ayush's route is 0.7 km or 700m longer than Sumit route.
Step-by-step explanation:
Ayush's route is 1km 2hm long while sumit route is 2hm 30dam.
We know that,
1 km = 10 hm
1 dam = 0.1 hm
Using these conversions we get
Ayush's route = 1km 2hm = (1×10) hm + 2 hm = 12 hm
Sumit route = 2hm 30dam = 2 hm + (30×0.1) hm = 2 hm + 3 hm = 5 hm
Ayush's route is longer.
Difference = 12 hm - 5 hm = 7 hm = 0.7 km [1 km = 10 hm]
Hence, Ayush's route is 0.7 km or 700 m longer than Sumit route.
Answer:
2 x^3
Step-by-step explanation:
sqrt(12 x^8)/ sqrt(3x^2)
combine into 1 term
sqrt(12x^8/3x^2)
when dividing exponents with the same base, we subtract
sqrt(12/3 x^(8-2))
sqrt(4 x^6)
sqrt(4) sqrt(x^6)
2 x^3
Answer
40.9
Step-by-step explanation:
Answer:
z= 3.63
z for significance level = 0.05 is ± 1.645
Step-by-step explanation:
Here p = 42% = 0.42
n= 500
We formulate our null and alternative hypotheses as
H0: p= 0.42 against Ha : p> 0.42 One tailed test
From this we can find q which is equal to 1-p= 1-0.42 = 0.58
Taking p`= 0.5
Now using the z test
z= p`- p/ √p(1-p)/n
Putting the values
z= 0.5- 0.42/ √0.42*0.58/500
z= 0.5- 0.42/ 0.0220
z= 3.63
For one tailed test the value of z for significance level = 0.05 is ± 1.645
Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.
