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3 years ago

120 dollars is how many pounds

Mathematics

1 answer:

inna [77]3 years ago

8 0

$120 USD is equivalent to £84.55

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How do i reflect something on a grid?

Xelga [282]

Answer:

Think of reflect as looking in a mirror. The question should tell you how your reflecting. It might say "reflect the rectangle over the y axis" (Look at the attachment below to see example).

Step-by-step explanation:

I can help more if I see the question. But usually questions ask you to reflect over the y or x axis.

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3 years ago

Which equation does not define a function?

stira [4]

The first relation gives two y-values for any given x-value (except x=6).

y² +x = 6 . . . . . is not a function

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3 years ago

arc GT = 18 degrees, arc TA =32 degrees, arc NA =112 degrees, arc IN = 74 degrees. find measure of angle GET

Svet_ta [14]

Answer:

Step-by-step explanation:

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3 years ago

Can someone please help with this math question:)

Bingel [31]

Solution:

The rationalisation factor for $\frac{1}{ a - \sqrt{b} }$ is $a + \sqrt{b}$

So, let us apply it here.

$\frac{1}{5 - \sqrt{2} }$

The rationalising factor for 5 - √2 is 5 + √2.

Therefore, multiplying and dividing by 5 + √2, we have

$= \frac{1}{5 - \sqrt{2} } \times \frac{5 + \sqrt{2} }{5 + \sqrt{2} } \\ = \frac{5 + \sqrt{2} }{(5 - \sqrt{2})(5 + \sqrt{2} ) } \\ = \frac{5 + \sqrt{2} }{ {(5)}^{2} - ( \sqrt{2})^{2} } \\ = \frac{5 + \sqrt{2} }{25 - 2} \\ = \frac{5 + \sqrt{2} }{23}$

Answer:

$\frac{5 + \sqrt{2} }{23}$

Hope you could understand.

If you have any query, feel free to ask.

5 0

2 years ago

Read 2 more answers

Y=arccos(1/x) Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago