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3 years ago

Domain Range

Mathematics

1 answer:

saul85 [17]3 years ago

6 0

No it's not.

There is no function that can produce this domain.

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Lyn asked some of her classmates how many people are normally at home for dinner. She recorded her results in the histogram show

erma4kov [3.2K]

Step-by-step explanation:

There is no diagram so I cannot understand how to answer

3 0

3 years ago

WILL GIVE BRAINLIEST ANSWER!

notka56 [123]

$f(x)=13|x-5|-6$ is the equation represented by all the operations.

<u>Step-by-step explanation:</u>

The given function is

$f(x)=|x|$

It is vertically compressed by a factor of 13 then the function becomes,

$f(x)=13|x|$

It shifted down 6 units, then the function becomes,

$\mathbf{f}(\mathbf{x})=\mathbf{1 3}|\boldsymbol{x}|-\boldsymbol{6}$

It shifted right by 5 units then the function becomes,

$f(x)=13|x-5|-6$

Thus the resultant equation thus obtained is $f(x)=13|x-5|-6$

6 0

3 years ago

big pictures are 56$ each small ones are 34$. 97 were sold at the total of 4,376. how many big and small pictures were sold

iVinArrow [24]

Answer:

The number of big pictures are 49 and number of small pictures are 48 .

Step-by-step explanation:

Let us assume that the number of big pictures are x .

Let us assume that the number of small pictures are y .

As given

Big pictures are 56$ each small ones are 34$.

97 were sold at the total of $4,376.

Equations becomes

x + y = 97

56x + 34y = 4376

Multiply x + y = 97 by 56 and subtracted from 56x + 34y = 4376 .

56x - 56x + 34y - 56y = 4376 - 5432

-22y = - 1056

22y = 1056

$y = \frac{1056}{22}$

y = 48

Put value of y in the equation x + y = 97

x + 48 =97

x = 97 - 48

x = 49

Therefore the number of big pictures are 49 and number of small pictures are 48 .

6 0

4 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago

What do i do on this question

Umnica [9.8K]

Find what the question mark is and both equation should have the same answers
(Im pretty sure)

5 1

3 years ago

Read 2 more answers