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jek_recluse [69]
3 years ago
15

Without finding the value of theta, please, if possible​

Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0
Hay mate!!!

please see the attachment...

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PLEASE HELP! ( 20 points! )
sattari [20]

Answer:

15.46

Step-by-step explanation:

3 0
3 years ago
If 32^2c= 8^c+7, what is the value of c
Leona [35]

\bf 32^{2c}=8^{c+7}~~ \begin{cases} 32=2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^5\\ 8=2\cdot 2\cdot 2\\ \qquad 2^3 \end{cases}\implies (2^5)^{2c}=(2^3)^{c+7} \\\\\\ 2^{5(2c)}=2^{3(c+7)}\implies \stackrel{\textit{if the bases are the same, then so are the exponents}}{5(2c)=3(c+7)\implies 10c=3c+21} \\\\\\ 7c=21\implies c=\cfrac{21}{7}\implies c=3

3 0
3 years ago
Read 2 more answers
If the population of a town grew 21% up to 15,049, what was the population last year?
katrin2010 [14]
15,049 x .21 = 3,160
subtract: 15,049 - 3,160 = 11, 888
3 0
3 years ago
Subtract 2-3/4-1 1/10=​
umka21 [38]

Answer:

23/20        

Step by step Explanation

5 0
3 years ago
Read 2 more answers
Please help. and please explain the process too
denis-greek [22]
The polynomial as per the given of the problem is f(x)=3x²-21x+30
Find the roots of this polynomial:
 x₁=[-b+√b²-4ac)]/2a and x₂ =[-b-√b²-4ac)]/2a
Plug in : x₁ = [21+√(21²-4(3)(30)]/6 = [21+√81}/6 ; x₁ = 5
and x₂ = [21-√(21²-4(3)(30)]/6 = [21-√81]/6 ; x₂ = 2
 Remark: I don't understand where did you bring √5 , it should be 5
6 0
4 years ago
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