Answer:
64√2 or 64 StartRoot 2 EndRoot
Step-by-step explanation:
A 45-45-90 traingle is a special traingle. Let's say one of the leg of the triangle is x. The other one is also x because of the isosocles triangle theorem. Therefore, using the pytagorean theorem, you find that x^2+x^2=c^2. 2(x)^2=c^2. You then square root both sides and get c= x√2.
Therefore, the two legs are x and the hypotenuse is x√2. x√2=128 because the question says that the hypotenuse is 128. Solve for x by dividing both sides by √2. X=128/√2. You rationalize it by multiplying the numberator and denominator of the fraction by √2. √2*√2= 2.
X=(128√2)/2= 64√2 cm.
Since X is the leg, the answer would be 64√2
Answer:
well that depends on how many pieces of gum are in one pack, once you have that info multiply it by 4.
Step-by-step explanation:
So, we know the perimeter is 24 cm. and the top part is the left part, (x) + 2, but remember, we need to have it be a number that, multiplied by 2, is 24. So, using this formula, the length, (top / bottom) is 7, and x = 5. Because 5 + 5 = 10, and when length is +2, it adds up to 14. 10 + 14 = 24. So, length is 7, width, (left, right) is 5.
Answer:
34.6 units
Step-by-step explanation:
The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.
The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).
Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:
Distance between point A(-6, 2) and point B(2, 6):

Let,





(nearest tenth)
Distance between B(2, 6) and C(7, 1):

Let,





(nearest tenth)
Distance between C(7, 1) and D(3, -5):

Let,





(nearest tenth)
Distance between D(3, -5) and A(-6, 2):

Let,





(nearest tenth)
Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units
D-value (meteorology) in meteorology refers to the deviation of actual altitude along a constant pressure surface from the standard atmosphere altitude of that surface.