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3 years ago

Can someone plz help me.

2 answers:

7 0

There should be 6 answers. TKE TEK KTE KET ETK EKT

All you have to do is start with one letter for example "T" and rotate the two after them until you run out of possibility like "TKE" "TEK" now if we were to rotate the letters after the first one again it would be "TKE" so we know that we are out of possibilities. Then you repeat with another letter.

hichkok12 [17]3 years ago

6 0

It will be 6 answers. TKE TEK KTE KET ETK EKT

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Rate of change from X = 15 to X = 19.

vova2212 [387]

Answer:

26%

Step-by-step explanation:

Subtract the old amt from the new amt, then divide by the old amt:

19-15/15 = 4/15 = 26.6%. Rounded to the nearest tenth it would be 26%.

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2 years ago

Can someone please help? what is the missing length?

Harrizon [31]

31

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3 years ago

Use the remainder theorem to find P (-3) for P (x) = x to the power of 4 + 4x to the power of 3 - 3x - 7

Nastasia [14]

Answer:

3, 4, 8

Step-by-step explanation:

5 0

2 years ago

Could anybody help with this confused ?!!

ss7ja [257]

X=4 y=9

4 [times] 4-9=7

3 0

3 years ago

Consider the function on the interval (0, 2π). f(x) = sin x + cos x (a) Find the open intervals on which the function is increas

Annette [7]

Answer:Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)

Step-by-step explanation:

given f(x) = sin(x) + cos(x)

f(x) can be rewritten as $\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} } ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\$

Using these result in equation a we get

f(x) = $\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)$

Now we know that for derivative with respect to dependent variable is positive for an increasing function

Differentiating b on both sides with respect to x we get

f '(x) = $f '(x)=\sqrt{2} \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0$

where x∈(0,2π)

we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]

Thus for cos(π/4+x)>0 we should have

1) π/4 + x < π/2 => x<π/4 => x∈[0,π/4]

2) π/4 + x > 3π/2 => x > 5π/4 => x∈[5π/4,2π]

from conditions 1 and 2 we have x∈(0,π/4)∪(5π/4,2π)

Thus the function is decreasing in x∈(π/4,5π/4)

5 0

2 years ago