If the lengths of two sides of a right triangle are 5 and 12 units, what is the least possible length, in units, of the third si
de? Express your answer in simplest radical form.
2 answers:
Answer:
13
Step-by-step explanation:
(leg)^2 + (leg)^2 = (hypotenuse)^2
(5)^2 + (12)^2 = (h)^2
25 + 144 = (h)^2
169 = (h)^2
sqrt{169} = sqrt{(h)^2}
-13 = h
13 = h
We reject h = -13 because length is a distance and distance cannot be negative.
So, h = 13 units is the answer.
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Use PEMDAS:
P Parentheses first
E Exponents (ie Powers and Square Roots, etc.)
MD Multiplication and Division (left-to-right)
AS Addition and Subtraction (left-to-right)
and 
Step-by-step explanation:
y = x² + x + 5
y = x + 1
x² + x + 5 = x + 1
x² + 4 = 0
x² = -4
x = ±√-4 undefined
No solutions
Option → B
Answer:
29/7
Step-by-step explanation:
4 1/7 = 4 + 1/7
4 = 28/7
then:
4 1/7 = 28/7 + 1/7 = (28+1) / 7
= 29/7
x-intercept(s):
(
−
1
2
,
0
)
Find the y-intercepts.
y-intercept(s):
(
0
,
2
3
)
List the intersections.
x-intercept(s):
(
−
1
2
,
0
)
y-intercept(s):
(
0
,
2
3
)
54.7 cm because a squared + b squared = c squared
