Answer:
d
Step-by-step explanation:
just makes sense bc they're all in 2nd grade
k = 13The smallest zero or root is x = -10
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Steps:
note: you can write "x^2" to mean "x squared"
f(x) = x^2+3x-10
f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5
f(x+5) = (x^2+10x+25)+3(x+5)-10
f(x+5) = x^2+10x+25+3x+15-10
f(x+5) = x^2+13x+30
Compare this with x^2+kx+30 and we see that k = 13
Factor and solve the equation below
x^2+13x+30 = 0
(x+10)(x+3) = 0
x+1 = 0 or x+3 = 0
x = -10 or x = -3
The smallest zero is x = -10 as its the left-most value on a number line.
Please mark brainliest
<em><u>Hope this helps.</u></em>
Answer:
See below.
Step-by-step explanation:
Fifth root of 243 = 3,
Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),
then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).
Equating equal parts and using de Moivre's theorem:
r^5 =243 and cos 5Ф + i sin 5Ф = cos 240 + i sin 240
r = 3 and 5Ф = 240 +360p so Ф = 48 + 72p
So Ф = 48, 120, 192, 264, 336 for 48 ≤ Ф < 360
So there are 5 distinct solutions given by:
3(cos 48 + i sin 48),
3(cos 120 + i sin 120),
3(cos 192 + i sin 192),
3(cos 264 + i sin 264),
3(cos 336 + i sin 336).. (Answer).
I think they are perpendicular to each other
