2 years ago

What is the range of this graph?

Mathematics

2 answers:

Nikolay [14]2 years ago

8 0

(-3,1] is the correct one

Sergio039 [100]2 years ago

4 0

Answer:

I think the answer is (-3,1)

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Which expressions are equivalent to 4x^4-9x^2?

weeeeeb [17]

We can cosider this to be a difference of 2 squares so

4x^4 - 9x^2 = (2x^2 - 3x)(2x^2 + 3x) so D is one answer

also we could take x^2 out and get
x^2(4x^2 - 9) = C

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2 years ago

Help fast 15 points

den301095 [7]

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1/4

Step-by-step explanation:

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2 years ago

49 more than the quotient of an unknown number and 12 is 53.

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2 years ago

Read 2 more answers

Really need help guys!<br> Love you if you can solve them wiz steps!!!!

Marta_Voda [28]

Answer:

Step-by-step explanation:

$x^2+y^2-8x+4y+4=0\\\\a)\\\\x^2-8x+y^2+4y+4=0\\\\x^2-2*x*4+(y^2+2*y*2+2^2)=0\\\\x^2-2*x*4+4^2+(y+2)^2=4^2\\\\(x-4)^2+(y+2)^2=4^2\\$

Hence,

The radius of the circle is 4 units, coordinates of its centre are (4,-2).

$b)\\\\y=0\\x^2+0^2-8x+4*0+4=0\\x^2-8x+4=0\\a=1\ \ \ \ b=-8\ \ \ \ c=4\\D=(-8)^2-4*1*4\\D=64-14\\D=48\\\sqrt{D}=\sqrt{48} \\ \sqrt{D}=\sqrt{16*3} \\\sqrt{D}=\sqrt{4^2*3} \\\sqrt{D}=4\sqrt{3} \\\displaystyle\\x=\frac{-(-8)б4\sqrt{3} }{2*1} \\\\x=\frac{8б4\sqrt{3} }{2} \\x_1=4-2\sqrt{3} \\x_2=4+2\sqrt{3}$

$c)\\\\A(6,2\sqrt{3}-2)\\ (x-4)^2+(y+2)^2=4^2\\(6-4)^2+(2\sqrt{3} -2+2)^2=16\\2^2+(2\sqrt{3} )^2=16\\4+2^2*(\sqrt{3})^2=16\\ 4+4*3=16\\4+12=16\\16\equiv16$

$d)\\A(6,2\sqrt{3}-2)}\\\sqrt{3} x+3y=\\\sqrt{3}(6)+3(2\sqrt{3} -2)=\\ 6\sqrt{3}+6\sqrt{3} -6=\\ 12\sqrt{3}-6$

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11 months ago