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3 years ago

What do you need to know in order to do a rotation? Complete a rotation of your choice and identify the rule for this

Mathematics

1 answer:

Maurinko [17]3 years ago

6 0

You are suppose to use a ruler to rotate the thing you need to rotate

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5) v= 2V A) Domain: { All real numbers. } Range: { All real numbers. } B) Domain: x s 2 Range: y20 C) Domain: x 20 Range: y20 D)

Eddi Din [679]

We will have the following:

*It will have a domain of x equal or greater than 0:

$x\ge0$

It will have a domain of y equal or greater than 0:

$y\ge0$

***

We have the following:

We determine the domain as follows:

Since the function is subject to a root, no negative value can be put in it, so its domain (Since x is not accompanied by any value in the root) can only "hold" values equal or greater than 0.

We determine the range as follows:

Since the function is subject to a root, it's solutions will never be negative, so its range can only hold values equal or greater than 0.

8 0

1 year ago

Sally has a GPA of 3.35. What percentage of the students have a GPA above her GPA?

vichka [17]

I'm hoping it might be around the .65 percentile?

3 0

3 years ago

An automobile originally purchased for $18,000 is losing value at a rate of 15% per year. The number of dollars it will be worth

matrenka [14]

Answer:

Step-by-step explanation:

8 0

3 years ago

mack plans to meet his 4 friends. how many different ways can he make his visit if he visits each friend once?

natima [27]

He can invite all his friends to the park or somewhere and meet them all at once

4 0

4 years ago

Please help need answers

Kaylis [27]

$\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet$

$\frak{Good\;Morning!!}$

$\pmb{\tt{Question\;1}}$

$\star\boldsymbol{\rm{Given-:}}$

Side length = 7 cm,
Side length = 5 cm.

$\star\boldsymbol{\rm{We're\;looking\;for-:}}$

Side length = x

This is how it's done.

$\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet$

There's a special formula that we can use if we need to find the longest side of a right triangle. Fortunately, all of these triangles are right ones! Good.

The formula is. $\boldsymbol{\rm{a^2+b^2=c^2}}$ . This formula is known as Pythagoras' Theorem. This formula only works for right triangles.

Since we have a and b, we can just put in the values (7 for a and 5 for b), And then simplify!

$\boldsymbol{\rm{7^2+5^2=c^2}}$ | 7^2 simplifies to 49, and 5^2 simplifies to 25

$\boldsymbol{\rm{49+25=c^2}}$ . | add

$\boldsymbol{\rm{74=c^2}}$ | square root both sides

$\boldsymbol{\rm{8.6=c}}}$ . | the answer is given to 1 decimal place, as the problem required

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$\pmb{\tt{Question\;2}}$

Once more, we're given two sides, and asked to find the third one,

which is still the longest side.

$\boldsymbol{\rm{a^2+b^2=c^2}}$ is still the formula used here

Put in 5 for a and 3 for b.

$\boldsymbol{\rm{5^2+3^2=c^2}}$ | 5^2 simplifies to 25, and 3^2 simplifies to 9

$\boldsymbol{\rm{25+9=c^2}}$ | add

$\boldsymbol{\rm{34=c^2}}$ | square root both sides

$\boldsymbol{\rm 5.8=c}}$ | once again it's given to one decimal place

$\hspace{350pt}\above5$

$\pmb{\tt{Question\;3}}$

This problem is solved the exact same way

$\boldsymbol{\rm{a^2+b^2=c^2}}$

$\boldsymbol{\rm{8.2^2+4.7^2=c^2}}$

$\boldsymbol{\rm{67.24+22.09=c^2}}$

$\boldsymbol{\rm{89.33=c^2}}$

$\boldsymbol{\rm{9.5=c}}$ , rounded to one D.P.

$\orange\hspace{300pt}\above2$

$\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet\equiv\bullet$

$\pmb{\tt{Question4}}$

Here we have the longest side and one side length-:

$\boldsymbol{\rm{4^2+b^2=7^2}}$ | 4^2 simplifies to 16 and 7^2 simplifies to 49

$\boldsymbol{\rm{16+b^2=49}}$ | subtract 16 from both sides

$\boldsymbol{b^2=33}$ | square root both sides

$\boldsymbol{\rm{b=5.7}}$

$\orange\hspace{300pt}\above3$

$\pmb{\tt{Question\;5}}$

$\boldsymbol{\rm{3.8^2+b^2=7.9^2}}$

$\boldsymbol{\rm{14.44+b^2=62.41}}$

$\boldsymbol{\rm{b^2=47.97}}$

$\boldsymbol{\rm{b=6.9}}$

$\orange\hspace{300pt}\above3$

$\pmb{\tt{Question\;6}}$

$\boldsymbol{\rm{a^2+6.1^2=7.3^2}}$

$\boldsymbol{\rm{a^2+37.21=53.29}}$

$\boldsymbol{\rm{a^2=16.08}}$

$\boldsymbol{\rm{a=4.0}}$

$\pmb{\tt{done~!!!}}$

$\orange\hspace{300pt}\above3$

7 0

2 years ago