Pythagoras Theorem:
hipotenuse²=leg₁²+leg₂²

First posible triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=5 (5²=25)

13³=144 + 25

Answer:can be side lengths of a triangle

Second triangle:
hypotenuse=12.6 (12.6²=158.76)
leg₁=6.7 ( 6.7²=44.89)
leg₂=6.5 (6.5²=42.25)

leg₁²+leg₂²=44.89+42.25=87.14≠158.76

Answer: cannot be side lenghts of a triangle.

third triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=11 (11²=121)

leg₁²+leg₂²=144+121=265≠169

Answer: cannot be side lenghts of a triangle.

fourth triangle:
hypotenuse=13 (13²=169)
leg₁=6 ( 6²=36)
leg₂=4 (4²=16)

leg₁²+leg₁²=36+16=52≠169

Answer: cannot be side lenghts of a triangle.

8 0

3 years ago

Read 2 more answers

Find the inverse of the function.<br> f(x) = -3x + 7

LUCKY_DIMON [66]

F(x) = -3x + 7
y = -3x + 7
x = -3y + 7
-3y + 7 = x
-3y = x - 7
y = -1/3x + 7/3
f^-1(x) = -1/3x + 7/3

8 0

2 years ago

NEED HELP ASAP WILL GIVE 100 PTS IN IT HAVE TO BE RIGHT AND I WILL GIVE BRAINIEST

AlladinOne [14]

Answer:

Step-by-step explanation:

(7^0/7^5)^1/5

= (1 / 7^5)^ 1/5

= (7^-5)^1/5

= 7^-1

= 1/7 is the correct answer.

5 0

2 years ago

Learning the basics 6.23 suppose you select one value from a uniform distribution with a=0and b=10.what is the probability that

IgorC [24]

The area under the graph of the continuous uniform distribution is 1.
a. The probability that the value will be between 5 and 7 is the area between 5 and 7.
$P(5 \ \textless \ x \ \textless \ 7) = \frac{2}{10} = \frac{1}{5}$
b. The probability that the value will be between 2 and 3 is the are between 2 and 3.
$P(2 \ \textless \ x \ \textless \ 3)=\frac{1}{10}$
c. The mean is given by:
$\mu=\frac{a+b}{2}=\frac{0+10}{2}=5$
d. The variance is given by:
$\sigma^2}=\frac{1}{12}(b-a)^{2}=\frac{1}{12}(10)^{2}=8.3333$
The standard deviation is:
$\sqrt{\sigma} = \sqrt{8.3333} =2.887$

5 0

3 years ago