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2 years ago

Zane calculated the missing side length of one of these triangles using the Pythagorean Theorem. Which triangle is it?

Mathematics

2 answers:

topjm [15]2 years ago

6 0

The pythagorean theorem states that for any RIGHT-ANGLED TRIANGLE, the relationship between the side lengths are A²+B²=C², where C is the hypotenuse :P

Which means that it can only be applied on right angled triangles~
So your answer would be B, a right-angled triangle.

Ask me if you are unsure about anything <3

Andrei [34K]2 years ago

4 0

It has to be triangle B.

You only use the pythagorean theorem for right triangles (triangles with one 90 degree angle).

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Enter the indicated element.

Ymorist [56]

$a_{1,3} = 9$

Step-by-step explanation:

The elements of matrices are represented in the form

$a_{r,c}$

Here r is the row number and c is the column number.

Given matrix is:

$A = \left[\begin{array}{ccc}3&6&9\\2&4&8\end{array}\right]$

The elements of matrix will be represented by a_{r,c}

We have to find $a_{1,3}$

Which means that the first row and third column.

The first row is: 3 6 9 and the element in third column is: 9

So,

$a_{1,3} = 9$

Keywords: Matrices, Elements

Learn more about matrices at:

#LearnwithBrainly

6 0

2 years ago

If log(x + y) = log 3 + ½log x + ½log y, prove that x² + y² = 7xy

Free_Kalibri [48]

Answer:

See Explanation

Step-by-step explanation:

$log(x + y) = log3 + \frac{1}{2} logx+ \frac{1}{2} logy \\ \\ log(x + y) = log3 + logx ^{\frac{1}{2}} + logy ^{\frac{1}{2}}\\ \\ log(x + y) = log3 + log(xy) ^{\frac{1}{2}} \\ \\ log(x + y) = log[3(xy) ^{\frac{1}{2}}] \\ \\ x + y = 3(xy) ^{\frac{1}{2}} \\ \\ squaring \: both \: sides \\ {(x + y)}^{2} = \bigg(3(xy) ^{\frac{1}{2}} \bigg)^{2} \\ \\ {x}^{2} + {y}^{2} + 2xy = 9xy \\ \\ {x}^{2} + {y}^{2} = 9xy - 2xy \\ \\ \purple{ \bold{{x}^{2} + {y}^{2} = 7xy}} \\ thus \: proved$

7 0

2 years ago

Read 2 more answers

Please help and thank you

d1i1m1o1n [39]

# ov wekz..... \_ >_○ _/

5 0

2 years ago

The velocity v of the flow of blood at a distance r from the central axis of an artery of radius R is given below, where k is th

andrew11 [14]

The average rate of blood flow is given by

$\displaystyle\frac1{R-0}\int_0^Rv(r)\,\mathrm dr=\frac kR\int_0^R(R^2-r^2)\,\mathrm dr$
$=\dfrac kR\left(R^2r-\dfrac13r^3\right)\bigg|_{r=0}^{r=R}$
$=\dfrac kR\left(R^3-\dfrac13R^3\right)$
$=\dfrac{2kR^2}3$

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2 years ago

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Lubov Fominskaja [6]

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