Answer:
X=6
Step-by-step explanation:
Answer: -5100
<u>Step-by-step explanation:</u>
![\sum^4_1[100(-4)^{n-1}]\qquad \rightarrow \qquad a_1=100\ \text{and r = -4}\\\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\\\S_4=\dfrac{100(1-(-4)^4)}{1-(-4)}\\\\\\.\quad=\dfrac{100(1-256)}{1+4}\\\\\\.\quad=\dfrac{100(-255)}{5}\\\\.\quad=20(-255)\\\\.\quad=-5100\\](https://tex.z-dn.net/?f=%5Csum%5E4_1%5B100%28-4%29%5E%7Bn-1%7D%5D%5Cqquad%20%5Crightarrow%20%5Cqquad%20a_1%3D100%5C%20%5Ctext%7Band%20r%20%3D%20-4%7D%5C%5C%5C%5C%5C%5CS_n%3D%5Cdfrac%7Ba_1%281-r%5En%29%7D%7B1-r%7D%5C%5C%5C%5C%5C%5C%5C%5CS_4%3D%5Cdfrac%7B100%281-%28-4%29%5E4%29%7D%7B1-%28-4%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%3D%5Cdfrac%7B100%281-256%29%7D%7B1%2B4%7D%5C%5C%5C%5C%5C%5C.%5Cquad%3D%5Cdfrac%7B100%28-255%29%7D%7B5%7D%5C%5C%5C%5C.%5Cquad%3D20%28-255%29%5C%5C%5C%5C.%5Cquad%3D-5100%5C%5C)
Let x and y be the 2 parts of 15 ==> x + y=15 (given)
Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)
Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)
==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy
But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50
Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
The answer is a) equilateral triangle. If you want to inscribe a hexagon inside a circle, the tools or constructions that should be used is 6 equilateral triangles. If you draw a hexagon inscribed in a circle and draw radii to the corners of the hexagon, you will create triangles, six of them.<span>
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If it were like that, 9/20 would be baseball card.
x=27 since 20x3=60
