<p>Hi,</p>
<p>I've completed all the questions in the BB, but there are three maths questions that I don't understand. They are-</p>
<p>Page 548: Question 16-</p>
<p>A cube with volume 8 cubic centimetres is inscribed in a sphere so that each vertex of the cube touches the sphere. What is the length of the diameter of the sphere?</p>
<p>The answer is 2 x 3^1/2 (2 times the square root of 3)</p>
<p>Page 457: Question 19-</p>
<p>The square of x is equal to four times the square of y. If x is one more than twice y, what is the value of x?</p>
<p>The answer choices are -
a) -4
b) -1/2
c) -1/4
d) 1/4
e) 1/2</p>
<p>I tried to enter the answers into the equation x^2 = 4y^2 but not of them seemed to fulfil the requirement (x 1 more than twice y)
The answer is E.</p>
<p>Page 642: Question 17-</p>
<p>In the xy-plane, line l passes through the origin and is perpendicular to the line 4x + y = k. If the two lines intersect at the point (t, t + 1), what is the value of t?</p>
<p>The answer choices are-
a) -4/3
b) -5/4
c) 3/4
d) 5/4
e) 4/3
The answer is A</p>
<p>Any help would be HUGELY appreciated.</p>
<p>Q16: Basically, you need to find the length of the diagonal between the farthest pair of vertices of the cube. Draw a 3d view of a cube and hopefully you’ll see this.
- first find the length of the diagonal of the side of the cube using pythagorian.
d(side) = sqrt(2^2 + 2^2) = 2 * sqrt(2)
- next find the length of the diagonal of the cube again with pythagorian.
d(cube) = sqrt(d(side)^2 + 2^2) = sqrt(12) = 2 * sqrt(3)</p>
<p>Q19: Solve a simple algebraic equations (2 equations with 2 unknowns) or using your method:
- substitute x=1/2 into x^2 = 4y^2, you’ll get y = {1/4, -1/4}
- substitute the resulting y into x = 2y+1, you’ll get x = {3/2, 1/2}
- the answer then is 1/2</p>
<p>Q17: The perpendicular line is y = 1/4 x. If the point is (t, t+1), then t = -4/3.
- Just substitute x=t and y=t+1 into y = 1/4 x and solve for t.</p>
<p>Thanks for all of those, they really helped!
Just one more that I forgot to put earlier- </p>
<p>Page 773 - Question 18:</p>
<p>h(t)= c-(d-4t)^2</p>
<p>At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height in feet, after t seconds was given by the function above, in which c and d are positive constants. If the ball reached its maximum height of 106 at time t=2.5, what was the height, in feet of the ball at time t=1?</p>
<p>I have just no idea on this one…</p>
<p>It is a problem where you need to set the equations up based on the wording given.</p>
<p>At time t=0: h(0) = c-(d-4<em>0)^2 = c-d^2 (and this is given as 6 feet)
At time t=2.5: h(2.5) = c-(d-4</em>2.5)^2 = c-(d-10)^2 (and this is given as 106)</p>
<p>So we again have 2 equations and 2 unknowns:</p>
<p>c-d^2 = 6 and c-(d-10)^2 = 106</p>
<p>Solve for c and d, and if I do my algebra right, c = 106 and d = 10.</p>
<p>Then h(1) = 106-(10-4*1)^2 = 106-36 = 70</p>
<p>You sir (or madam) are a legend.</p>
<p>Huge gratitude sent your way.</p>
<p>In Question 19, if you attempt to solve for the 2 equation system with 2 unknowns you end up with:</p>
<p>X = X+1</p>
<p>How can you go from there to X = 1/2?</p>
<p>Thanks</p>