Math Genius?? Help me out! :] thx?

Question

1. What is the smallest value of x if 12÷(x+1) is an integer?

2. A right rectangular prism has a base area one quarter the base area of a larger right rectangular prism. The height of the smaller prism is half the height of the larger one. What is the ratio of the volume of the smaller prism to the volume of the larger prism? express answer as a common fraction.

3. In triangle ABC the measure of angle A is x degrees, the measure of angle C is 5x degrees. what is the value of x? express your answer as a decimal to the nearest tenth.

4. What is the sum of the even positive integers less than 62?

5. A right triangle is inscribed in a circle with a diameter of 100 units long. What is the maximum area of the triangle in square units??


thx you guys!! <33

Answer 1

I'll do the 1st 4 and most of no. 5.
1. We know that x+1 must divide 12 and the smallest
possible integer that can work is x = -13.
If x = -13, x+1 = -12, the smallest divisor of 12.

2. Volume of large prism = lwh, area of base = lw.
Volume of smaller prism = 1/4*lw *1/2*h = 1/8*lwh
Ratio of volumes = 1/8.

3. Sum of angles = 180 degrees.
Problem is angle B = 180-6x and there is
no way to find x from this equation. Something's
missing here!

4. Sum of even integers less than 62 = sum of
even integers <= 60 = sum of all integers <= 60 -
sum of odd integers <= 60.
Sum of all integers <= 60 = 60*61/2 = 30*61 = 1830.
Sum of odd integers less than 60 .
There are 30 odd integers less than 60
and their sum is 30² = 900.
Sum of all even integers less than 62 = 930.
(Here integers means "positive integers".)

5. Let a and b be the lengths of the legs. First, the
right angle must be inscribed in a semicircle, so
the hypotenuse must be a diameter of the circle.
Then a²+ b² = 10000. The area of the triangle is ½ab.
So we want to maximise ½ab subject to the
constraint a² + b² = 10000.
So b = √(10000-a²).
So what is the maximum of A =½a√(10000-a²)
on the interval [0,100]?
This is a problem in elementary calculus.
You take the first derivative of A with respect to a,
set it to 0 and solve for a.
If you do this you will find that a = √5000 and
b also = √5000.
So the maximum area is ½(5000) = 2500.

Answer 2

1. The true answer to this question is -13, since 12÷(12-13) = 12/-12 or -1 -- an integer. Using only positive integers, the answer would be 1, since 12÷ (1 + 1) = 2. If x can be 0, the answer is 0, since 12/12 = 1. Answer -13 for this, and show your deep thinking.

2. the ratio is 1/8 as a common fraction. This can be shown, since B x H = 1/4 x 1/2 = 1/8.

3. This question is incomplete. You need to have angle B in terms of x as well to solve this. All you can say is B = 180 - 6x at this point.

4. This sum is 930. This requires finding the sum of the 30 numbers 2,4,6,...,60. This is : (First number + Last number) times (half the count of 30) -- or (2 + 60)(15) or 930

5. The answer is 2,500 square units. To find this one, realize that the diameter of the circle is the hypotenuse of the right triangle. So a² + b² = 100²
Since you want to maximize the area, you get the lengths a and b as close together as possible. Making them equal is as close as you can get. So we now have a = b, and substitute b for a in the formula to get 2a² = 100². Multiply it out to get 2a² = 10,000, and a² = 5,000. But since a = b, this is also ab = 5000. Plug this into the formula for the area of a triangle and get A = 1/2(5000) = 2500

Answer 3

1. -13
2. The formula for volume is V=Bh for right prisms, and so if we quadruple the base area B and double the height h, we wil multiply V by 8 (octuple?)
3. Not enough information is provided to answer this question.
4.
2 + 4 + 6 + 8 + ... + 60
= (2+60) + (4+58) + (6+56) + ... + (30+32)
= 15 * 62
= 930
5. If a right triangle is inscribed in a circle, the hypotenuse must be a diameter (from the Inscribed Angle Theorem). So we only need to maximize the height. But this is easy, we just use an isoceles right triangle with height equal to the radius of the circle. So the base of the triangle measures 100 units and the height 50 units, giving an area of 2500 square units.