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Need hellp with 4 math questions. 10 easy points!?
1) Write the coordinates of the point of intersection of the graph of 2x-3y+8=0 and the x-axis. I think the graph is invalid because on my TI-84+ silver edition it comes out invalid. My calc might just be buggin but... See what you get.
2) If x=4, the value of 4x^1/2 +(x^0+3)^-1
3) solve for y: (sqrt y-3)= 5-y
4) The graph of the equation 2x^2+y^2=3 is
A) circle B) ellipse c) hyperbola d) parabola
Justify your answer.
step by step please?
1. The equation isn't in the proper form...I'm not sure if it's supposed to be quadractic or linear so....My guess would be linear. It comes up as invalid, because x and y shouldn't be on the same side. It should be y= whatever to graph it properly.
So if 2x-3y+8=0 then y= 2/3x + 3/8
The line intersects the x-axis at (-4,0)
2. 4x^1/2 +(x^0+3)^-1
If x=4 then, 4(4^1/2)+ (4^0+3)^-1
Simplify the numbers inside the parethesis.
4(2)+(1+3)^-1
And finish solving..
8+4^-1
8+.25
y=8.25
3. (sqrt y-3)= 5-y
Square both sides.
y-3=y^2-10y+25
Set it equal to zero
y^2-11y+28=0
Solve using the quadratic formula, which I'm not going to type out but you probably know it. Or for this one, you can just search for two factor of 28 and add up to -11. For this one it's simply -4 and -7. You divide/multiply those both by -1.
So Y= 4 or 7
4. Ellipse, because the equation isn't in the proper form to be any of the others.
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