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AlexFokin [52]
3 years ago
8

What is the formula for the area of a triangle?

Mathematics
1 answer:
SashulF [63]3 years ago
6 0

<em>The</em><em> </em><em>formula</em><em> </em><em>to</em><em> </em><em>find</em><em> </em><em>the</em><em> </em><em>area</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>triangle</em><em> </em><em>is</em><em> </em><em>:</em>

<em>1</em><em>/</em><em>2</em><em> </em><em>base</em><em>*</em><em>height</em>

<em>For</em><em> </em><em>inst</em><em>ance</em><em>:</em>

<em>base</em><em>=</em><em>4</em><em> </em><em>cm</em>

<em>height</em><em>=</em><em>5</em><em>c</em><em>m</em>

<em>area</em><em> </em><em>of</em><em> </em><em>traingle</em><em>=</em><em>1</em><em>/</em><em>2</em><em>*</em><em>b</em><em>*</em><em>h</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>1</em><em>/</em><em>2</em><em>*</em><em>4</em><em>*</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>1</em><em>0</em><em>c</em><em>m</em><em>^</em><em>2</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

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3 years ago
Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11)
Katen [24]

Answer:

\frac{y^{2} }{81} -\frac{x^{2} }{40} }=1 is standard equation of hyperbola with vertices at (0, ±9) and foci at (0, ±11).

Step-by-step explanation:

We have given the vertices at (0, ±9) and foci at (0, ±11).

Let (0,±a)  = (0,±9) and (0,±c)  = (0,±11)

The standard equation of parabola is:

\frac{y^{2} }{a^{2} } -\frac{x^{2} }{b^{2} }=1

From statement, a  = 9

c² = a²+b²

(11)²  = (9)²+b²

121-81  = b²

40  = b²

Putting the value of a² and b² in standard equation of parabola, we have

\frac{y^{2} }{81} -\frac{x^{2} }{40} }=1 which is the answer.

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There could be a strong correlation between the proximity of the holiday season and the number of people who buy in the shopping centers.

It is known that when there are vacations people tend to frequent shopping centers more often than when they are busy with work or school.

Therefore, the proximity in the holiday season is related to the increase in the number of people who buy in the shopping centers.

This means that there is a strong correlation between both variables, since when one increases the other also does. This type of correlation is called positive. When, on the contrary, the increase of one variable causes the decrease of another variable, it is said that there is a negative correlation.

There are several coefficients that measure the degree of correlation (strong or weak), adapted to the nature of the data. The best known is the 'r' coefficient of Pearson correlation

A correlation is strong when the change in a variable x produces a significant change in a variable 'y'. In this case, the correlation coefficient r approaches | 1 |.

When the correlation between two variables is weak, the change of one causes a very slight and difficult to perceive change in the other variable. In this case, the correlation coefficient approaches zero

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kogti [31]
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