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Inga [223]
2 years ago
10

Skip count by twos. 30,_ _ _ _ _ _ _ _ _,50.

Mathematics
1 answer:
Vadim26 [7]2 years ago
5 0
30,32,34,36,38,40,42,44,46,48,50
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Measurement of <1 please help
GrogVix [38]

Answer:

27°

Step-by-step explanation:

Triangles when all three sides are added up make up 180°. So In this problem you would need to add up 80 and 73 (153) then subtracting that number by 180 (27) to find the value/measurement of <1.

4 0
2 years ago
Which inequality is graphed in the coordinate plane
Alborosie

it is B

im not sure what else to say but my answer has to be 20 words or something

7 0
3 years ago
Pls help i need this question by 8:00am tomorrow
kolezko [41]

Answer:

the slope is 5

explanation:

goes up 5 and right 1

4 0
3 years ago
Read 2 more answers
Xe^(-x^2/128) absolute max and absolute min
adoni [48]
f(x)=xe^{-x^2/128}
\implies f'(x)=e^{-x^2/128}+x\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}

Extrema can occur when the derivative is zero or undefined.

\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}=0\implies 1-\dfrac1{64}x^2=0\implies x^2=64\implies x=\pm8

Maxima occur where the first derivative is zero and the second derivative is negative; minima where the second derivative is positive. You have

f''(x)=-\dfrac1{32}xe^{-x^2/128}+\left(1-\dfrac1{64}x^2\right)\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(-\dfrac3{64}x+\dfrac1{4096}x^3\right)e^{-x^2/128}

At the critical points, you get

f''(-8)=\dfrac1{4\sqrt e}>0
f''(8)=-\dfrac1{4\sqrt e}

So you have a minimum at \left(-8,-\dfrac8{\sqrt e}\right) and a maximum at \left(8,\dfrac8{\sqrt e}\right).

Meanwhile, as x\to\pm\infty, it's clear that f(x)\to0, so these extrema are absolute on the function's domain.
6 0
3 years ago
In need of assistance
Mkey [24]

Answer:

60

Step-by-step explanation:

_---&$'_-4$':/-&$$--(:'##4--&&4___&$$_666&4&&&&&

7 0
1 year ago
Read 2 more answers
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