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3 years ago

15

Is the initial graph a function ? Is the inverse of the graph a function ?

1 answer:

sertanlavr [38]3 years ago

7 0

Answer:

a) No

b) Yes

Step-by-step explanation:

a) The value Temp = 2 maps to more than one value of Time, so Time is NOT a function of Temp.

b) Each value of Time only maps to one value of Temp, so Temp IS a function of Time.

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If x-y = 40 and x+y = 70, then what is the value of x squared -y squared ?

OlgaM077 [116]

Let's see what to do buddy...

____________________________

Reminder:

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

So we need to just Multiply above equations like this :

$x + y = 70$

$x - y = 40$

$(x + y)(x - y) = 70 \times 40$

${x}^{2} - {y}^{2} = 2800$

____________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

7 0

3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

What's the answer to this problem

ryzh [129]

B, because it is the one that is closest together, and it is like a upside down V, unlike Graph A, which is spread out so much.

5 0

3 years ago

Fifteen percent of 168

allsm [11]

$\large\text{Hey there!}$

$\rm{15\%\ of\ 168}\\\\\rm{= 15\% \times 168} \\\\\rm{= \dfrac{15}{100}\times 168}\\\\\rm{= \dfrac{15}{100}\times \dfrac{168}{1}}\\\\\rm{=\dfrac{15\times168}{100\times1}}\\\\\rm{= \dfrac{2,520}{100}}\\\\\rm{= 25.2}$

$\huge\text{Therefore, your answer should be: }$

$\huge\boxed{\frak{25.2}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

8 0

1 year ago

Read 2 more answers

Justin Bieber's sand castle is 180 cm tall but the wind is

larisa [96]

Answer:

Smh why justin bieber tho.

Step-by-step explanation:

8 0

3 years ago