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3 years ago

There are values of t so that sin t = 5/4. True or false.

Mathematics

2 answers:

Romashka [77]3 years ago

3 0

False sin t can never be more then 1

AnnyKZ [126]3 years ago

3 0

Answer: The given statement is FALSE.

Step-by-step explanation: We are given to check whether the following statement is TRUE of FALSE :

"There are values of t so that $\sin t=\dfrac{5}{4}.$ "

We know that the range of sine function is the close interval [-1, 1].

That is,

the value of sine of any angle cannot be less than -1 and greater than 1.

According to the given information, we have

$\sin t=\dfrac{5}{4}=1.25>1.$

So, no such values of t exists for which $\sin t=\dfrac{5}{4}.$

Hence, the given statement is FALSE.

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Read 2 more answers

The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on

babymother [125]

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Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

Let the length of the poster be x and the width be y.

Given that the area of the poster is 96 cm².

∴xy =96

$\Rightarrow y= \frac{96}{x}$

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

= (x-6) cm

The width of the printing space is =(y-2.2) cm

=( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)

$\Rightarrow A=(x-6)(\frac{96}{x}-4)$ [ Putting $y= \frac{96}{x}$ ]

$\Rightarrow A=120-\frac{576}{x}-4x$

Differentiating with respect to x

$A'= \frac{576}{x^2}-4$

Again differentiating with respect to x

$A''=-\frac{1152}{x^3}$

To find the minimum area, we set A'=0

$\therefore \frac{576}{x^2}-4=0$

$\Rightarrow \frac{576}{x^2}=4$

$\Rightarrow x^2=\frac{576}{4}$

$\Rightarrow x^2 =144$

$\Rightarrow x=\pm 12$

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

$\therefore A''|_{x=12}=-\frac{1152}{12^3}$

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is $y=\frac{96}{12}$ = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

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3 years ago