2^a = 5^b = 20^c, express c in terms of and b

c=ab/(2b+a)

Step-by-step explanation:

20^c = 4^c  *  5^c

20^c=  (2^2)^c   * 5^c

20^c= (2 )^(2c)   * ( 5 )^c

Raise both sides to power a

20^(ca)=(2^a)^(2c)  * (5  )^(ac)

20^(ca)=(20^c)^(2c) * (5  )^(ac)

Raise both sides to power b

20^(cab)=(20)^(2c^2b)*(5^b)^(ac)

20^(cab)=20^(2c^2b) * (20^c)^(ac)

20^(cab)=20^(2c^2b) * 20^(ac^2)

Rewriting using law of exponents on right hand side

20^(cab)=20^(2c^2b+ac^2)

Now bases are same so that means the exponents have to be the same, that is we have:

cab=2c^2b+ac^2

assuming c is not 0, divide by c on both sides

ab=2cb+ac

Factor the right hand side

ab=c(2b+a)

Divide both sides by (2b+a)

c=ab/(2b+a)

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°