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algol [13]
3 years ago
10

Can 4,9, 15 represent the lengths of the sides of a triangle?

Mathematics
2 answers:
REY [17]3 years ago
6 0

Answer:

No

Step-by-step explanation:

By the converse of Pythagoras' identity.

If the longest side squared equals the sum of the squares of the other 2 sides then triangle is right.

longest side = 15

15² = 225

4² + 9² = 16 + 81 = 97 ≠ 225

Thus sides do not represent a right triangle.

Gwar [14]3 years ago
4 0

Answer:

No

Step-by-step explanation:

4,4,5 and 12,7,7 are isosceles triangles because they have two equal sides . 15,15,15 is an equilateral triangle -it has three equal sides

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(40/5)-7+2? I know the answer but I don’t know how I got to it
ExtremeBDS [4]

Answer:

3

Step-by-step explanation:

(40/5)-7+2

PEMDAS says parentheses first, so divide inside the parentheses

(8)-7+2

Then add and subtract from left to right

1 +2

3

8 0
3 years ago
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Step-by-step explanation:

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natima [27]

Answer:

A table that has (0,0), (-1,1), (-4, 2) and undefined for any positive x value

Step-by-step explanation:

Reflecting across the y axis just changes the x values, it makes them negative.  so \sqrt{x} has points (0,0), (1,1), (4, 2) and so on.  reflecting over the y axis makes them(0,0), (-1,1), (-4, 2) and again so on.

Also good to mention in \sqrt{x} negative x values are undefined, so flipped over the y axis positive x values are undefined.

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6 0
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Norma-Jean [14]

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Step-by-step explanation:

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7 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

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3 years ago
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