My Space Mathematician Teacher needs this answer,

1 answer:

4 0

219864070186 - 277409297400 = -57545227214 x 780498 = -4.4913935e+16 - 1274806769367926946610431 = -1.2748068e+24 x 3.6614422e+21 = -4.6676314e+45 - 437913 = -4.6676314e+45 - 88878 = -4.6676314e+45

Ф = 500 , 5,000

5,000 / ∞ x ∞ x ∞ = 5,000^+++∞/ 68307147 = 0.00007319878^+++∞ - 917 - 974 - 9 - 19344381 = -19346280.9999^+++∞

-4.6676314e+45 -19346280.9999^+++∞ = -4.6676314e+45^+++∞

I believe that will be your answer.

I tried my best.

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Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor

Alex787 [66]

The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

the lengths of the sides of BRAD are:

$\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$

The lengths of the sides of KELY are:

$\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}$

Each side of KELY is twice the length of the corresponding side on BRAD. This makes the ratio of the sides 2:1 and the figures are similar.

8 0

2 years ago

Read 2 more answers

Find cos q if q is an angle in standard position and the point with coordinates (3, –4) lies on the terminal side of the angle.

gayaneshka [121]

Answer:

cos q = 3/5

Step-by-step explanation:

Standard position means the vertex (point or corner of the angle) is at (0,0) and one side of the angle is glued to the positive x-axis (facts, but not technical math terms) See image. Special triangles have all three sides nice and clean with whole number lengths, we call these Pythagorean triples. 3-4-5 is your most basic Pythagorean triple. So we don't even have to calculate the hypotenuse, see image. Now the triangle shown is easy to work with, using entry-level trig...cos = ADJ/HYP. So we get 3/5=.6 BUUuuuut, the angle q in the original problem is actually the giant angle, marked in yellow (see image) and we're in the fourth quadrant which means there's negative numbers all over the place. So just to be sure the answer is .6 and not -.6 Check your signs. One trick to remember is the ASTC markings in the quadrants. I use All Students Take Calculus, but what it means is in the first quadrant All the trig functions are positive. Only Sine (and fam) are positive in the 2nd quadrant. Tan (and fam) in the 3rd and Cos and fam in the 4th quadrant. It's a good quick check.

cos q = 3/5 OR cos q = .6