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dangina [55]
3 years ago
5

3. Determine whether the set of numbers can be the measures of the sides of a triangle. If so, classify the triangle as acute, o

btuse, or right. 16, 30, 33
Mathematics
2 answers:
mihalych1998 [28]3 years ago
6 0

Answer:

yes. it will be an acute.

julia-pushkina [17]3 years ago
3 0

Answer:

all of them are acute

.......

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A girl obtained 32.8 marks in nepali and failed by 7.2 marks. What was the pass marks of the subject?
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40 marks was the pass marks of the subject.

Step-by-step explanation:

32.8 + 7.2 = 40 marks         { basic addition theory based }

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How will the graph of line y = -5x + 7 be affected if the 7 is changed to –7?
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Answer:a

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What is the z-score for a patient who takes 7 days to recover? (choose the closest possibility.)
Kisachek [45]

Based on the time it took the patient to recover and the mean and standard deviation, the z-score is 1.5.

<h3>What is the z-score?</h3>

The z-score for this patient who takes 7 days to recover can be found as:

= (Time taken to recover - Mean) / Standard deviation

Solving for the z-score gives:

= (7 - 5.2) / 1.2

= 1.8 / 1.2

= 1.5

The first part of the question is:

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.2 days and a standard deviation of 1.2 days.

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5 0
2 years ago
Find the linearization of f(x) = x ^ (y / 2) a = 43
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DWAFECGSAD

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7 0
3 years ago
I don't know if this is right... please someone help mee
worty [1.4K]
For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
6 0
3 years ago
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