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stiks02 [169]
3 years ago
8

Help with this math question WILL GIVE BRAINLIEST

Mathematics
1 answer:
sergeinik [125]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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Two altitudes of a triangle have lengths $12$ and $14$. What is the longest possible integer length of the third altitude
nataly862011 [7]

The longest possible altitude of the third altitude (if it is a positive integer) is 83.

According to statement

Let h is the length of third altitude

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.

From Area of triangle

A = 1/2*B*H

Substitute the values in it

A = 1/2*a*12

a = 2A / 12 -(1)

Then

A = 1/2*b*14

b = 2A / 14 -(2)

Then

A = 1/2*c*h

c = 2A / h -(3)

Now, we will use the triangle inequalities:

  • a < b+c

2A/12 < 2A/14 + 2A/h

Solve it and get

h<84

  • b < a+c

2A/14 < 2A/12 + 2A/h

Solve it and get

h > -84

  • c < a+b

2A/h < 2A/12 + 2A/14

Solve it and get

h > 6.46

From all the three inequalities we get:

6.46<h<84

So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.

Learn more about TRIANGLE here brainly.com/question/2217700

#SPJ4

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2 years ago
FRAME AN EQUATION AND SOLVE IT. NEED HELP ASAP I only have 5 minutes!​
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Answer:

<em>The answer to your question is</em><em> 125</em>

Step-by-step explanation:

<em> x = 25 X 5= 125.</em>

<em>The required number is 125.</em>

<u><em>I hope this helps and have a good day!</em></u>

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2 years ago
Can someone please help me and show work too. If you can thank you
djyliett [7]

Answer:

3n-7

Step-by-step explanation:

8 0
2 years ago
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