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Cerrena [4.2K]
3 years ago
6

Next two sets of numbers after 3-6, 7-14

Mathematics
1 answer:
kicyunya [14]3 years ago
8 0
Next stop sets are 15-30 and 31-62
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Solve for x. 5^11x=25^x+9
4vir4ik [10]
   
\displaystyle\\
5^{11x}=25^{x+9}\\\\
5^{11x}=\Big(5^2\Big)^{x+9}\\\\
5^{11x}=5^{2(x+9)}\\\\
11x = 2(x+9)\\\\
11x = 2x+18\\\\
11x - 2x = 18\\\\
9x = 18\\\\
x =  \frac{18}{9} = \boxed{2}



6 0
3 years ago
Will mark as brainlist please
valina [46]

Answer:

maximum

vertex at (-1,1)

axis of symm:  x = -1

2 solutions

(-2,0) and (0,0)

Step-by-step explanation:

6 0
3 years ago
Evaluate the function. h(n)=n with the power of 3 -4 find h(-2)
Alisiya [41]

Answer:

stuff with math

Step-by-step explanation:

8 0
3 years ago
What is the answer to 2+b=7
Natasha_Volkova [10]

Answer:

b=5

Step-by-step explanation:

2+b=7

Subtract 2 from each side

2+b-2=7-2

b= 5

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%28sinx%5E%7B2%7D%20theta%29%5Cfrac%7Bx%7D%7By%7D%281%2Bcostheata%29" id="TexFormula1" title="
beks73 [17]

The result of expanding the trigonometry expression \sin^2(\theta) * (1 + \cos(\theta)) is cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

<h3>How to evaluate the expression?</h3>

The expression is given as:

\sin^2(\theta) * (1 + \cos(\theta))

Express \sin^2(\theta) as 1 - \cos^2(\theta).

So, we have:

\sin^2(\theta) * (1 + \cos(\theta)) =  (1- \cos^2(\theta)) * (1 + \cos(\theta))

Open the bracket

\sin^2(\theta) * (1 + \cos(\theta)) =  1 + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Express 1 as cos°(Ф)

\sin^2(\theta) * (1 + \cos(\theta)) =  cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Hence, the result of expanding the trigonometry expression \sin^2(\theta) * (1 + \cos(\theta)) is cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Read more about trigonometry expressions at:

brainly.com/question/8120556

#SPJ1

3 0
2 years ago
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