<p>if f(x) =ax^2 +bx +c, how must a and b be related so that the graph of f(x-3) will be symmetric about the y-axis?</p>
<p>a = b/6 ? (10 char)</p>
<p>and the circle is x^2 + y^2 = 4</p>
<p>How did you solve it? Please post the process</p>
<p>3x-4y = 10
-4y= -3x + 10
y= (3/4)x - (10/4)</p>
<p>The line crosses the X axis at (0,3) and the Y axis at (2,0)
Since the line is tangent, the circle can’t include (0,3).</p>
<p>So…</p>
<p>equation of circle : (x-h)^2+(y-k)^2=r^2
so answer is</p>
<p>x^2+y^2=4</p>
<p>I don’t feel like typing out the work lol.</p>
<p>For the circle, I know that a line that goes through the center of the circle is perpendicular to the tangent line. So i found that line and found its intersection with the tangent line. Then distance between the origin and that point and squared it.</p>
<p>For the the second one, symmetry across the y-axis means it is an even function. That means f(x) = f(-x). Set f(x) equal to f(x-3) and solve for a. The c’s cancel and the last x’s cancel.</p>
<p>f f(x) =ax^2 +bx +c, how must a and b be related so that the graph of f(x-3) will be symmetric about the y-axis?</p>
<p>IlDesi has answered correctly… Just use f(x-3) = f( -(x-3) ) for symmetry across y axis, and you’ll get the required condition(s).</p>
<p>What’s the equation of the circle that has its center at the origin and is tangent to the line with equation 3x-4y = 10?</p>
<p>Distance of a point (x1,y1) from a line ax + by + c = 0 is
ax1 + by1 + c / sqrt(a^2 + b^2)</p>
<p>So here , distance between 4y+10-3x = 0, and the point (0,0) is equal to the radius of the circle and = 4(0) - 3(0) + 10 / 5
= 2</p>
<p>So the eqn. is x^2 + y^2 = 2^2</p>
<p>Actually I kind of enjoy wasting a googolplex of spiders’ time :)</p>
<p>Hopefully people try the problems before looking at the answer. So …</p>
<p>The sum of the first 10 terms of an arithmetic sequence is 100. The sum of the first 100 terms is 10. The sum of the first 110 terms is:</p>
<p>A) 90
B) -90
C) 110
D) -110
E) -100</p>
<p>Sum of n terms = n/2 [ 2a + (n-1)d]</p>
<p>100 = 5 [ 2a + 9d]
20 = 2a + 9d</p>
<p>10 = 50[ 2a + 99d]
10 = 50 [ 20 + 90d]
1 = 100 + 450d
d = -11/50</p>
<p>20 = 2a + 9d
20 + 99/50 = 2a
1099/100 = a</p>
<p>Sum of 110 terms = 55 [ 2(1099/100) + 109 (-11/50)]
= -110</p>
<p>Lol, I think I’m gonna just solve tough genuine questions from now on, or you’re gonna eat up my time :P</p>
<p>These questions are worded somewhat poorly, but the answers provided are correct. For those of you who are confused, the first question should read “less than EVERY other element in the set.” If it was less than any other element in the set, it would be a very different question.</p>
<p>If data is multiplied by a factor of b, the new std deviation = b * old std deviation. Adding or subtracting only changes the mean.</p>
<p>The weirdest that usually comes up is contrapositive. Just know that “If x then y” has the same truth value as " If not y then not x." Ex: If I go to the store then I will buy food has the same truth value as If I do not buy food then I do not go to the store.</p>
<p>Try this one .. it’s cool. I made it up :).Two legs of a right triangle have lengths cos (2a+b+213c) and cos(90-2a-b-213c). 0<2a+b+213c<90. (Sorry I can’t put the degree symbol in, but you get what I mean)What is the length of the hypotenuse?
a) 3/2+cos(a^2/b)+4c/5
b) 5sin(2abc) + 1/2
c) 5cos(a^(bc)) + 3/4
d) 1
e) None of the above</p>
<p>anyone? Hint: 90-2a-b-213c = 90 - (2a+b+213c)</p>
<p>Square of Length of hypotenuse = Sum of squares of other 2 sides
= cos ^2 (2a+b+213c) + cos ^2 (90-(2a+b±213c))
= cos ^2 (2a+b+213c) + sin ^2 (2a+b+213c)
= 1<br>
(because sin^2 @ + cos ^2 @ = 1)</p>
<p>So, hypotenuse is of length 1 unit.</p>
<p>are all of these questions included in SAT II Math I or Math II…?</p>
<p>Lol, the questions on the SAT math1 and 2 are much easier than this. Don’t worry.</p>
<p>
</p>
<p>Since when is 0 positive?</p>