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3 years ago

15

Determine whether the relation is a function. (-3,3), (-2,2), (-1,1), (1,-1), (2,-2), (3,-3)

1 answer:

Luba_88 [7]3 years ago

3 0

Answer:

Function

Step-by-step explanation:

Because there is an output for every input listed of its opposite.

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A builder needs 6.5 cubic metres of concrete for a job. This table shows the mixture for the concrete.

TiliK225 [7]

If the four parts sand means 4 cubic meters of sand per 13 cubic meters of concrete, you can you can multiply 4/13 by 6.5 to get 2 cubic meters of sand.

Let me know if I misinterpreted the question so that I can try to fix it. I hope this helps

6 0

4 years ago

Find the inverse of the given function.

Bogdan [553]

Answer:

The inverse is

<h2> $y =\sqrt{\frac{(\frac{x-3}{3}) ^{2}+5}{2}}$ </h2>

Step-by-step explanation:

$y=3\sqrt{2x^{2} -5} + 3$

To find the inverse of the function interchange the terms that's x becomes y and y becomes x

We have

$x=3\sqrt{2y^{2}-5 }+3$

Now solve for y

<u>Move 3 to the other side of the equation</u>

$3\sqrt{2y^{2}-5 } = x-3$

Divide both sides by 3

We have

$\sqrt{2y^{2}-5 } =\frac{x-3}{3}$

<u>square both sides of the equation to remove the square root</u>

That's

$2y^{2}-5 = (\frac{x-3}{3}) ^{2}$

<u>Move 5 to the other side of the equation</u>

$2y^{2} = (\frac{x-3}{3}) ^{2}+5$

<u>Divide both sides by 2</u>

We have

$y^{2} = \frac{(\frac{x-3}{3}) ^{2}+5}{2}$

<u>Find the square root of both sides</u>

We have the final answer as

$y =\sqrt{\frac{(\frac{x-3}{3}) ^{2}+5}{2}}$

Hope this helps you

7 0

4 years ago

What is the vertical asymptote of the function?<br> O x=-10<br> O x=-2<br> O x = 0<br> Ox=2

Alexxx [7]

Answer:d

Step-by-step explanation:

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=Given%20%5C%3A%20cotA%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B3%7D%7D" id="TexFormula1" title="Giv

ruslelena [56]

<h3>Given :</h3>

$\tt cotA = \sqrt{ \dfrac{1}{3}}$

$\tt \implies cotA = \dfrac{1}{\sqrt{3}}$

<h3>To Find :</h3>

All other trigonometric ratios, which are :

sinA
cosA
tanA
cosecA
secA

<h3>Solution :</h3>

Let's make a diagram of right angled triangle ABC.

Now, From point A,

AC = Hypotenuse

BC = Perpendicular

AB = Base

$\tt We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}$

$\tt We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}$

$\tt \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}$

$\tt \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}$

$\tt \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)$

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

$\tt \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}$

$\tt \implies AC^{2} = 1x^{2} + 3x^{2}$

$\tt \implies AC^{2} = 4x^{2}$

$\tt \implies AC = \sqrt{4x^{2}}$

$\tt \implies AC = 2x$

Now,

$\tt sin \theta = \dfrac{perpendicular}{hypotenuse}$

$\tt \implies sinA = \dfrac{BC}{AC}$

$\tt \implies sinA = \dfrac{\sqrt{3}x}{2x}$

$\tt \implies sinA = \dfrac{\sqrt{3}}{2}$

$\Large \boxed{\tt sinA = \dfrac{\sqrt{3}}{2}}$

$\tt cos \theta = \dfrac{base}{hypotenuse}$

$\tt \implies cosA = \dfrac{AB}{AC}$

$\tt \implies cosA = \dfrac{1x}{2x}$

$\tt \implies cosA = \dfrac{1}{2}$

$\Large \boxed{\tt cosA = \dfrac{1}{2}}$

$\tt tan \theta = \dfrac{perpendicular}{base}$

$\tt \implies tanA = \dfrac{BC}{AB}$

$\tt \implies tanA = \dfrac{\sqrt{3}x}{1x}$

$\tt \implies tanA = \sqrt{3}$

$\Large \boxed{\tt tanA = \sqrt{3}}$

$\tt cosec \theta = \dfrac{hypotenuse}{perpendicular}$

$\tt \implies cosecA = \dfrac{AC}{BC}$

$\tt \implies cosecA = \dfrac{2x}{\sqrt{3}x}$

$\tt \implies cosecA = \dfrac{2}{\sqrt{3}}$

$\Large \boxed{\tt cosecA = \dfrac{2}{\sqrt{3}}}$

$\tt sec \theta = \dfrac{hypotenuse}{base}$

$\tt \implies secA = \dfrac{AC}{AB}$

$\tt \implies secA = \dfrac{2x}{1x}$

$\tt \implies secA = 2$

$\Large \boxed{\tt secA = 2}$

5 0

3 years ago

Read 2 more answers

A school found that the number of students buying lunch from the cafeteria had declined. The school wants to revise its current

vodka [1.7K]

Answer:

C

Step-by-step explanation:

I say the anwer is C because If the a box will be place in front of the school entrance for parents to drop off their meals.

5 0

3 years ago