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Kaylis [27]
3 years ago
6

Jakie is 5 years older than his brother kevin kevin is k years old write an algebraic expression

Mathematics
1 answer:
Grace [21]3 years ago
5 0
The algebraic expression could be K=5+k
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Xy+(4(20))>x-5y(2+9-7)
prohojiy [21]

I'll solve for y xy+(4(20))>x-5y(2+9-7) 

xy+4(20)>x-5y(2+9-7) 

xy+4(20)>x-5y(4) 

xy+4(20)>x-20y 

xy+80>x-20y 

xy+20y+80>x 

y(x+20)>-80+x 

y>(-80+x)/(x+20)


3 0
2 years ago
Use this image to classify each pair of lines.
irga5000 [103]

Answer:

Parallel

Step-by-step explanation:

7 0
2 years ago
Evaluate log1∕6 36. please help
Sholpan [36]

\log_{\tfrac 16}   36 \\\\\\=\dfrac{\ln 36}{ \ln \left(\tfrac 16\right)}\\\\\\=\dfrac{ \ln 6^2}{\ln (6^{-1})}\\\\\\=\dfrac{2 \ln 6}{-1 \ln 6}\\\\= - 2

4 0
2 years ago
Factorize this expression completely <br>using (a-b)² formula<br>3a²- 36a + 108​
Novosadov [1.4K]

Answer:

3a^2 - 36a + 108  = (\sqrt{3}a - 6\sqrt{3} )^2

Step-by-step explanation:

3a^2 = (\sqrt{3} a)^2\\108 = 9 \times 4 \times 3= 3^2 \times 2^2 \times 3 = (3\times2 \sqrt{3} )^2 = (6\sqrt{3})^2

so, a = \sqrt{3}  a \ , \ b = 6\sqrt{3} \\\\(a - b)^2 = a^2 -2ab + b^2\\\\-36a = -2ab\\\\-36a = -2 \times \sqrt{3} a \times 6\sqrt{3}

Therefore ,

3a^2 - 36a + 108 = (\sqrt{3}a )^2 - 2(\sqrt{3}a \times 6\sqrt{3} ) + (6\sqrt{3} )^2  = (\sqrt{3}a - 6\sqrt{3} )^2

6 0
3 years ago
Express (5a to the power 2) x (2b to the power 2) x (c) as a power of a product.
ella [17]

Answer:

50 a^2 b c^2

Step-by-step explanation:

I'm quite confused about your question, but I try my best. I have attached the explanation above. Hopefully this will help

4 0
2 years ago
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