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olga55 [171]
4 years ago
12

How do you solve −8≤2/5(k−2)

Mathematics
1 answer:
Ulleksa [173]4 years ago
8 0

−8≤2/5(k−2)

Multiply the numbers in the parenthesis by 2/5

-8\leq \frac{2}{5}k-\frac{4}{5}

Multiply both sides by 5

-40\leq 2k-4

Move the terms and change the sign

-2k\leq -4+40

Divide both sides by -2

-2k\leq 36

Final answer would be

k\geq -18

FYI you change the >/<  signs whenever you're dividing or multiplying a negative

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What is the inverse of y=pi/2+sin x​
Alekssandra [29.7K]

Answer:

f^{-1}(x)=\sin ^{-1}(x-\frac{\pi}{2})

Step-by-step explanation:

The given function is

y=\frac{\pi}{2}+\sin x

To find the inverse of this function, we interchange x and y.

x=\frac{\pi}{2}+\sin y

we now solve for y.

x-\frac{\pi}{2}=\sin y

Take the sine inverse of both sides to obtain;

\sin ^{-1}(x-\frac{\pi}{2})=y

Hence the inverse of the given function is;

f^{-1}(x)=\sin ^{-1}(x-\frac{\pi}{2})

where \frac{\pi}{2}-1\le x\le \frac{\pi}{2}+1

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3 years ago
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 <span><span><span>22x</span>−11 over x to the second power - x - 30.</span></span>
3 0
4 years ago
Choose ALL possible answers
muminat
Select all the correct answers:
1) Yes

2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)

3) Yes

4) No

5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20

rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18

rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15

rg=20>18=rf
rg=20>15=rh

6)  No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)
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a/b = c/d where ad = bc

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