Okay this equation really says is what is 30% of 248.

So, lets convert 30% to a fraction, 3/10 which is easier to work with.

All you have to do now is get out a calculator and do 248 *3/10 (or .3) and get 74.4

So subtract 74.4 and get

173.6

8 0

3 years ago

The number of E.coli bacteria cells in a pond of stagnant water can be represented by the function below, where A represents the

liq [111]

Answer:

<u>Option B. 59%</u>

Explanation:

The function that represents the number of E.coli bacteria cells per 100 mL of water as the time t years elapses is:

$A(t)=136(1.123)^{4t}$

The base, 1.123, represents the multiplicative constant rate of change of the function, so you just must substitute 1 for t in the power part of the function:

$rate=(1.123)^{4t}=(1.123)^4=1.590$

Then, the multiplicative rate of change is 1.590, which means that every year the number of E.coli bacteria cells per 100 mL of water increases by a factor of 1.590, and that is 1.59 - 1 = 0.590 or 59% increase.

8 0

4 years ago

If f(x-3)=x^2-4x+5, what's f(1-x)

olganol [36]

Answer:

$\large\boxed{f(1-x)=x^2-4x+5}$

Step-by-step explanation:

$f(x-3)=x^2-4x+5\\\\f((x-3)+3)\to\text{put x + 3 to the equation of the function:}\\\\f(x-3+3)=(x+3)^2-4(x+3)+5\\\text{use the distributive property and}\ (a+b)^2=a^2+2ab+b^2\\f(x)=x^2+2(x)(3)+3^2+(-4)(x)+(-4)(3)+5\\f(x)=x^2+6x+9-4x-12+5\qquad\text{combine like terms}\\f(x)=x^2+(6x-4x)+(9-12+5)\\f(x)=x^2+2x+2\\\\f(1-x)\to\text{put 1 - x to the equation of the funtion f(x):}\\\\f(1-x)=(1-x)^2+2(1-x)+2\\\text{use the distributive property and}\ (a-b)^2=a^2-2ab+b^2$

$f(1-x)=1^2-2(1)(x)+x^2+(2)(1)+(2)(-x)+2\\f(1-x)=1-2x+x^2+2-2x+2\qquad\text{combine like terms}\\f(1-x)=x^2+(-2x-2x)+(1+2+2)\\f(1-x)=x^2-4x+5$

4 0

3 years ago