Simplify the expression: (ab)³ (3a)²b

If human height were quantized in one-foot increments, what would happen to the height of a child as she grows up?

agasfer [191]

I believe what the problem means is that the height of the child, if he/she grows in height, the mesurements are equal to 1 ft, 2 ft, 3 ft, 4 ft, etc only. and no decimals or fractions in between. it is also said that a human stops growing at the age of 20 naturally, hence with this increment, a child can grow incredibly very tall.

7 0

3 years ago

PLEASE SHOW YOUR WORK!!!<br><br><br><br><br><br> Solve for y: 3^3 + 14 y- (25y - 13) = y + 7y - 9y

Marina86 [1]

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3^3+14*y-(25*y-13)-(y+7*y-9*y)=0

Equation at the end of step 1

((3³ + 14y) - (25y - 13)) - -y = 0

Pull out like factors :

40 - 10y = -10 • (y - 4)

Equation at the end of step3:

-10 • (y - 4) = 0

STEP4:

Equations which are never true:

Solve : -10 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

Solve : y-4 = 0

Add 4 to both sides of the equation :

y = 4

6 0

3 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

<em><u>Solution:</u></em>

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

<em><u>Use the following log rules</u></em>

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$