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2 years ago

This one is for CATNSIE!!!!! You got this! answer quick!!!!

Mathematics

1 answer:

CaHeK987 [17]2 years ago

4 0

Answer:

OH TY LOL

Step-by-step explanation:

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PLEASE HELP!!!

elena-s [515]

Coordinates after 270 clockwise rotation is (-6,-2), (-3,-5) and (-8, -5).

<h3>What is graph?</h3>

Graph is a mathematical representation of a network and it describes the relationship between lines and points.

Given coordinates are: (-2,6) , (-5, 3) and (-5, 8)

For 270° clockwise about the origin, we have the following rule:

(x, y)= (-y, x)

So,

coordinates (-2,6) change into (-6,-2)

coordinates (-5, 3) change into (-3,-5)

coordinates (-5, 8) change into (-8, -5).

Learn more about this concept here:

brainly.com/question/12673036

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7 0

1 year ago

Plssss help meeee!!!

pogonyaev

The answer is c b/c you multiply

3 0

2 years ago

Read 2 more answers

Find the volume of the cone.<br> Either enter an exact answer in terms of pi or use 3.14 for pi

Amanda [17]

Answer:

Volume is 20.93333333 units^3

Step-by-step explanation:

I used 3.14 for pi

Look at the attachment

3 0

2 years ago

Read 2 more answers

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

Can someone help me plz

FrozenT [24]

Answer:

I cant see it?

Step-by-step explanation:

3 0

2 years ago