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4 years ago

Verify seca(1-sina)(seca+tana)=1

Mathematics

2 answers:

Daniel [21]4 years ago

7 0

That is an interesting question
Here is how to reach it....

Refer to the pic I attached
.

I hope that helps!

beks73 [17]4 years ago

5 0

Seca(1-sina)(seca+tana)=1
Left hand side
=(1-sina) (seca +tana) / cosa
=seca +tana - sinaseca - sinatan / cosa
=seca + tana -tana -(sin^2a/cosa) / cosa
=(1/cosa - sin^2a/cosa) / cosa
= (1-sin^2a / cosa) / cosa
= (1-sin^2a)/ (cos^2a)
=1 (verified)

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Select all of the following tables which represent y as a function of x.x 5 10 11y 3 9 15x 5 10 11y 3 9 9x 5 10 10y 3 9 15

11Alexandr11 [23.1K]

Answer:

Table 1 and 2 represent a function

Step-by-step explanation:

Given

Table 1

x 5 10 11

y 3 9 15

Table 2

x 5 10 11

y 3 9 9

Table 3

x 5 10 10

y 3 9 15

Required

Determine which of the tables represent that y is a function of x

For a relation to be a function; the x values must be unique.

In other words, each x value must not be repeated;

Having said that;

Analyzing Table 1

Table 1

x 5 10 11

y 3 9 15

Note that the x rows are unique as no value were repeated;

Hence, Table 1 is a function

Table 2

x 5 10 11

y 3 9 9

Note that the x rows are unique as no value were repeated;

Hence, Table 2 is a function

Table 3

x 5 10 10

y 3 9 15

Note that the x rows are not unique because 10 was repeated twice;

Hence, Table 3 is not a function

7 0

3 years ago

Suppose that six individuals are interested in taking part in a study relating BMI to a number of health outcomes. The following

posledela

Answer:

a) Mean = 27.65

Median = 27.645

b) Relative Frequency = 33.33%

Step-by-step explanation:

We are given the following data set:

25.78, 21.06, 36.54, 29.51, 18.96, 34.05

a) Mean and Median

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{165.9}{6} = 27.65$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$