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pentagon [3]
3 years ago
8

What is the are a square with a side length of 3x + 4.

Mathematics
1 answer:
geniusboy [140]3 years ago
3 0
(3x + 14) (3x + 14) = 9x^2 + 24x + 16
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What is the inverse of the function y=x+5/7 ?
Zinaida [17]
X= y+5/7
7x=y+5
y= 7x-5
8 0
4 years ago
Find the term to make the trinomial into a perfect square x2+10/3+?
Alborosie

Answer:

25/9

Step-by-step explanation:

x^2 + 10/3 x

Take half of the x-term coefficient and square it.

1/2 * 10/3 = 10/6 = 5/3

(5/3)^3 = 25/9

Answer: 25/9

6 0
3 years ago
PLZ HELP!!! I Will give brainliest. What is the value of x in sin(3x)=cos(6x) if x is in the interval of 0≤x≤π/2
sertanlavr [38]

Answer:

sin(2x)=cos(π2−2x)

So:

cos(π2−2x)=cos(3x)

Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that

(π2−2x)=±3x

i)

π2=5x

x=π10

ii)

π2=−x

x=−π2

Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)

So we see that

(π2−2x−2π)=±3x

iii)

π2−2π=5x

x=−310π

iv)

π2−2π=−x

x=2π−π2=32π

Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)

So we get ∀n∈N

i) x=π10+2πn

ii) x=−π2+2πn

iii) x=−310π+2πn

(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)

So we conclude that there are really three solutions and then the periodic extensions of those three solutions.

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Related Questions (More Answers Below)

5 0
3 years ago
F(x)=2x-3 and g(x)=x^2-5, find (fog) (x) and (gof) (x)
Alexxandr [17]
F(g(x)) = 2(x^2-5)-3
= 2x^2-13

g(f(x)) = (2x-3)^2 - 5
= 4x^2 + 9 -12x -5
= 4x^2 -12x + 4
6 0
3 years ago
PLEASE HELP What input value produces the same output value for the two functions
I am Lyosha [343]

Sorry it's an exam. I wish that I could.

I sad I could not help you :(


6 0
3 years ago
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