From what I'm understanding of these questions, the biggest thing you need to answer these is the formulas for cylinders and triangular prisms. I'm not sure what the quantities are for either question so I'm going to work with made up numbers to give examples for the formulas. For number 2 with the cylinder, let's consider the formula first:
π × r2 × h <em>OR </em>pi (3.14) times radius squared times height
If you have the height and you have pi, all you need to take is the doubled radius (aka multiply it by 2) and plug that back into the formula. For the sake of an example, I'm going to make up the number 2 for the radius and 6 for the height. Here's what that would look like:
r = 2; double it, resulting in 4
pi x 4^2 x 6
3.14 x 16 x 6
= 301.44
Work with the actual numbers you have and you're good to go.
For number 3, reducing something by 1/2 means dividing by 2. Let's consider the formula and then work through another example:
1/2 x b x h x l <em>OR </em> 1/2 times base times height times length
For the sake of an example, I'll use 10 for the height, 15 for the base, and 20 for the length:
h = 10; reduce by 1/2, resulting in 5
1/2 x 15 x 5 x 20
= 750
Plug in your actual quantities, and remember your volume units. Hope this helps!
The perimeter of a parallelgram is the sum of the lengths of its four sides.
Parallelogram ABCD has sides AB, BC, CD, and AB.
Sides AB and CD are parallel and of equal length = 19 units.
Sides BC and CD are parallel and of equal length. Assuming thi is the length of 5 units given in the statement, the perimeter of the parallelogram ABCD is: 19 units + 19 units + 5 units + 5 units = 48 units.
Please, inform if the length of 5 units corresponds to other distance, but even in that case, with this explanation you should be able to calculate the perimeter of this and other parallelograms.
Answer: 48 untis.
Answer: B
Step-by-step explanation: The question is asking for the point greatly different from all the rest, the outlier. Point B is far away from the grouping of all the other points, making it the outlier.
Step-by-step explanation:
(1) Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial) 2x²+6x-20=0
2x²+6x-20
2(x²+3x-10)
the factors are 2 and (x²+3x-10)
(2) Factor the polynomial completely. (4 points: 2 point for each factor)
2(x²+3x-10)
2(x²-2x+5x-10)
2(x(x-2) + 5(x-2)) group like terms
2(x+5)(x-2)
(3) If a product is equal to zero, we know at least one of the factors must be zero. And the constant factor cannot be zero. So set each binomial factor equal to 0 and solve for x, the width of your project. (2 points: 1 point for each factor)
constant = 2 cannot be zero
the other factors are (x+5) and (x-2)
(x+5)=0 => x= -5
or
(x-2)=0 => x=2
(4) What are the dimensions of your project? Remember that the width of your project is represented by x. (2 points: 1 point for each dimension)
thank you so much, sorry if it's a little confusing!!
(it is indeed confusing, because physical dimensions cannot be negative)
The dimensions of the project (assumed a rectangle) are +2 and -5
Use substitution
-x + 4 = 1/3x + 8
4 = 4/3x + 8
-4 = 4/3x
x = -4 * 3/4
x = -3
Then plug x into one of the original equations. In this case y = -x + 4 is the easiest.
y = -(-3) + 4
y = 7
