If P is the orthocenter of △ABC, AB = 13, BF = 9,and FC = 5.6, find each measure & the perimeter of triangle ABC.
Answers
Answer:
[tex]Perimeter = 38.6[/tex]
Step-by-step explanation:
Given
[tex]A\ B = 13[/tex]
[tex]B\ F = 9[/tex]
[tex]F\ C = 5.6[/tex]
Required
Calculate the perimeter of A B C
To calculate the perimeter of A B C, we sum up the sides of the triangle
i.e.
[tex]Perimeter = A B + B\ C + A\ C[/tex]
AB is given in the question already.
So, we need to calculate B F and F C
To solve for BF, we consider triangle B F C
Using Pythagoras Theorem, we have:
[tex]B\ C^2 = B\ F^2 + F\ C^2[/tex]
Substitute values for BF and FC;
[tex]B\ C^2 = 9^2 + 5.6^2[/tex]
[tex]B\ C^2 = 81 + 31.36[/tex]
[tex]B\ C^2 = 112.36[/tex]
[tex]B\ C^2= \sqrt{112.36[/tex]
[tex]B\ C = 10.6[/tex]
Next, we calculate A C.
To calculate A C, we need to calculate A F, considering triangle B F A
Using Pythagoras Theorem, we have:
[tex]A\ B^2 = B\ F^2 + A\ F^2[/tex]
Substitute values for B F and A B;
[tex]13^2 = 9^2 + A\ F^2[/tex]
[tex]169= 81 + A\ F^2[/tex]
[tex]A\ F^2 = 169 - 81[/tex]
[tex]A\ F^2 = 88[/tex]
[tex]A\ F = \sqrt{88[/tex]
[tex]A\ F = 9.4[/tex]
Since F lies on line A C:
[tex]A\ C = A\ F + F\ C[/tex]
[tex]A\ C = 9.4 + 5.6[/tex]
[tex]A\ C = 15.0[/tex]
Recall that:
[tex]Perimeter = A\ B + B\ C + A\ C[/tex]
[tex]Perimeter = 13 + 10.6 + 15.0[/tex]
[tex]Perimeter = 38.6[/tex]
Hence, the perimeter is 38.6
Given that P is the orthocenter of triangle ABC, applying the properties of an orthocenter of a triangle and the Pythagorean Theorem, each measure and the perimeter of △ABC are:
AB = 13BC = 10.6AC = 15Perimeter = 38.6Recall:
An altitude is the line that connects the vertex of a triangle to the opposite side of which it is perpendicular to that side.Orthocenter is the point where all three altitudes intersect each other.Given that P is the orthocenter of triangle ABC, where,
AB = 13BF = 9FC = 5.6Perimeter of △ABC = AB + BC + AC
First, we need to find BC and AC.
△BFC is a right triangle, apply Pythagorean theorem to find BC:
[tex]BC = \sqrt{BF^2 + FC^2}[/tex]
Substitute[tex]BC = \sqrt{9^2 + 5.6^2}\\\\\mathbf{BC = 10.6}[/tex]
△BFA is a right triangle, apply Pythagorean theorem to find FA:
[tex]FA = \sqrt{AB^2 - BF^2}[/tex]
Substitute[tex]FA = \sqrt{13^2 - 9^2}\\\\\mathbf{FA = 9.4}[/tex]
AC = FA + FC
SubstituteAC = 9.4 + 5.6
AC = 15
Perimeter of △ABC = AB + BC + AC
SubstitutePerimeter of △ABC = 13 + 10.6 + 15
Perimeter of △ABC = 38.6 units
Therefore, given that P is the orthocenter of triangle ABC, applying the properties of an orthocenter of a triangle and the Pythagorean Theorem, each measure and the perimeter of △ABC are:
AB = 13BC = 10.6AC = 15Perimeter = 38.6Learn more here:
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Answers
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Answer:
Step-by-step explanation:
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Answers
Answer:
40 termsStep-by-step explanation:
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Required Answer:-first term =a=9Last term =l=165 Sn=3480As we know that in a AP
[tex]{:}\longrightarrow[/tex][tex]\sf S_n={\dfrac {1}{2}}(a+l)n[/tex]
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[tex]{:}\longrightarrow[/tex][tex]\sf 3480={\dfrac {1}{2}}(174)n[/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 3480={\dfrac {174}{2}}n [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 3480=87n [/tex]
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The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second). Find the probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)
Answers
Answer:
0.1827
Step-by-step explanation:
Given mean of exponential distribution = 100
==> 1/χ = 100 ==> χ = 1/100 ==> χ = 0.01
PDF of χ , f(x) = χe^(-χx), x ≥ 0
===> f(x) = 0.01e^(-0.01x), x ≥ 0
Now we find the probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day
P(X>170) = ∞∫170 f(x)dx
P(X>170) = ∞∫170 0.01 e^(0.01x) dx
P(X>170) = [e^(-0.01x) / -0.01]^∞ base 170
P(X>170) = -1 [e^-∞ - e^-0.01*170]
P(X>170) = e^-1.7
P(X>170) = 0.1827
The probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day is 0.1827
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Answers
Answer:
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Step-by-step explanation:
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What are there fractions equivalent to 6/24
Answers
Answer: Remember how to find equivalent fractions? Here are some:
1/4, 2/8, 3/12, 4/16, 5/20, 25/100, 90/360, 1800/7200
To find all of these, there's a simple method- multiply the denominator and the numerator by the same number. As long as you make sure to use the same number for both the top AND the bottom, you'll always have an equivalent fraction. For example,
1 1800 1800
--- x -------------- = ---------
4 1800 7200
This works because, as you may have noticed, any muliplier you use is in the form of n over n, which, of course, is equivalent to one whole or 100%. Make sense?
The three fractions equivalent to 6/24 are 3/12, 1/4, and 2/8.
What is a fraction?A fraction is a part of the whole represented by a/b, where a and b are any integers.
The given fraction is 6/24.
Simplify the fraction:
6/24
= 3/12
= 1/4
Divide and multiply the above fraction by 2:
2/8
Hence, the three fractions equivalent to 6/24 are 3/12, 1/4, and 2/8.
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