Answer:
y=0. 24x + 12. 7
Step-by-step explanation:
since you start with 12. 7 and gain .24 each year, you would multiply the years by. 24 and just add the 12. 7 I hope this makes sense. also I love your profile picture :)
Answer:
it should be 12
Step-by-step explanation:
I hope this help
<span>7√7 - 2√28
2 sqrt(28) = 4 * sqrt(7)
7*sqrt(7) - 4*sqrt(7) =
3 * sqrt(7)
</span>
Answer:
D) 53 Degrees.
Step-by-step explanation:
Things we need to establish beforehand: We know that Lines OZ and OX are equal because they are both radii of the circle. We can make an Iscoceles traingle by drawing a line between ZX. We know angle YZO and angle YXO is a right angle because YZ and XY are tangent to the circle. The Arc angle is the same angle as angle ZOX.
1) Find angles OZX and OXZ. these will be 26.5, because 180-127 is 53, which is the sum of the two angles. the two angles are the same, so divide 53 by 2.
2) Find Angles XZY and ZXY. We know that YZO is a right angle, and both XZY and OZX make up this right angle so XZY + OZX = 90. OZX is 26.5, so 90-26.5=XZY. XZY = ZXY, so both angles equal 63.5.
3) Now that we have two angles of triangle XYZ, we can find angle XYZ. 180-(XZY+ZXY)=XYZ, so (180-(63.5+63.5)=53. Angle XYZ=53.
the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.
multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.
so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5 <----- notice, current coefficient times 3 divided by 4+1.
anyhow, with that out of the way, lemme proceed in this one.

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.
so then, we know that the expanded 2nd term is 24x therefore

we also know that the expanded 3rd term is 240x², therefore we can say that

but but but, we know what "n" equals to, recall above, so let's do some quick substitution
![\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20a%5E2n%5E2-a%5E2n%3D480%5Cqquad%20%5Cboxed%7Bn%3D%5Ccfrac%7B24%7D%7Ba%7D%7D%5Cqquad%20a%5E2%5Cleft%28%20%5Ccfrac%7B24%7D%7Ba%7D%20%5Cright%29%5E2-a%5E2%5Cleft%28%20%5Ccfrac%7B24%7D%7Ba%7D%20%5Cright%29%3D480%20%5C%5C%5C%5C%5C%5C%20a%5E2%5Ccdot%20%5Ccfrac%7B24%5E2%7D%7Ba%5E2%7D-24a%3D480%5Cimplies%2024%5E2-24a%3D480%5Cimplies%20576-24a%3D480%20%5C%5C%5C%5C%5C%5C%20-24a%3D-96%5Cimplies%20a%3D%5Ccfrac%7B-96%7D%7B-24%7D%5Cimplies%20%5Cblacktriangleright%20a%20%3D%204%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20n%3D%5Ccfrac%7B24%7D%7Ba%7D%5Cimplies%20n%3D%5Ccfrac%7B24%7D%7B4%7D%5Cimplies%20%5Cblacktriangleright%20n%3D6%20%5Cblacktriangleleft)