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3 years ago

Tables do not include numerical data. True or false?

Mathematics

2 answers:

vovikov84 [41]3 years ago

6 0

Answer:

The answer to this statement is False.

Step-by-step explanation:

In mathematics table are a useful way for the representation and organization of the data in rows and columns or we can say the data points.

with the help of tables we may also plot our graph.

for example- the frequency table represents the numerical data in which the data is divided into two columns one of the interval and the other of the frequencies.

Hence, the statement:

Tables do not include numerical data is a False statement.

stepan [7]3 years ago

5 0

True all table does not need to include numerical data

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Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side

lorasvet [3.4K]

Answer:

Step-by-step explanation:

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

=cot x (sec²x)²

=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]

= cot x(1+ 2 tan²x +tan⁴x)

=cot x+ 2 cot x tan²x+cot x tan⁴x

=cot x +2 tan x + tan³x [ ∵cot x tan x $=\frac{ \textrm{tan x }}{\textrm{tan x}}$ =1]

=R.H.S

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

L.H.S =(sin x)(tan x cos x - cot x cos x)

= sin x tan x cos x - sin x cot x cos x

$=\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}$

= sin²x -cos²x

=1-cos²x-cos²x

=1-2 cos²x

=R.H.S

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

= $1+\frac{{sin^2x}}{cos^2x}$ [ $\textrm{sec x}=\frac{1}{\textrm{cos x}}$ ]

=1+tan²x $[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]$

=sec²x

=R.H.S

$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}$

L.H.S= $\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}$

$=\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}$

$=\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}$

$=\frac{\textrm{2sin x}}{sin^2 x}$

= 2 csc x

= R.H.S

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

= sec²x-tan²x

= $\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}$

$=\frac{1- sin^2x}{cos^2x}$

$=\frac{cos^2x}{cos^2x}$

8 0

3 years ago

What are the domain and range of this function y=(x+3)^2 -5

n200080 [17]

Answer: Choice A

Note: the range should be $[-5, \infty)$ . See explanation below.

===================================================

We can plug in any real number for x to get some output for y. The domain is the set of all real numbers in which we say $(-\infty, \infty)$ which is interval notation. It represents the interval from negative infinity to positive infinity.

The range is the set of possible outputs. The smallest output possible is y = -5 which occurs at the vertex (3,-5). We can get this y value or larger. So we can describe the range as the set of y values such that $y \ge -5$ and that translates to the interval notation $[-5, \infty)$ .

The square bracket says "include this endpoint" while the curved parenthesis says to exclude the endpoint. Your teacher mistakenly wrote $(-5, \infty)$ for choice A, when they should have written $[-5, \infty)$