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Nikolay [14]
3 years ago
12

The two shorter sides of a triangle are the same length. The length of the longer side is 5 m longer than each of the shorter si

des. The perimeter of the triangle is 29 m. Write and solve an equation to determine the length of the longest side of the triangle. Explain each step you perform.

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
6 0
Here's my answer! :)

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- 3x + y = 3<br> 2y = -3x - 12
Lynna [10]

Answer:

x=-2, y=-3

Step-by-step explanation:

we can rearrange the questions to make it easier

y=3x+3

2y=-3x-12

if you add the two equations, you get

3y=0x-9

so y=-9/3 = -3

plug y into one of the equations

-3=3x+3

-6=3x

x=-2

and then check your work by plugging into the other equation:

2*-3=-3(-2)-12

-6=6-12

-6=-6

8 0
3 years ago
What is the simplified form of StartRoot 10,000 x Superscript 64 Baseline EndRoot ?
Nat2105 [25]

Answer:

5000x^8

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Express the interval (-9,3]
aleksklad [387]
3-(-9)=12
answer of this question is 12
5 0
3 years ago
The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are
raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

A_{i}: the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

P(A_{i})=0.1 for i = 1, 2, 3, 4, 5

The complement for A_{i} is given by

A_{i}^{$c$}: the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P(A_{i}^{$c$})=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=(0.9)^{5}=0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P(A_{1}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5})=5×(0.1)×(0.9)^{4}=0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P(A_{1}∪A_{2}∪A_{3}∪A_{4}∪A_{5})=1-P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=1-0.59049=0.40951

6 0
3 years ago
A b c or d??? create a direct variation equation ​
iren [92.7K]

Step-by-step explanation:

Y= -5x would be your answer

4 0
4 years ago
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