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3 years ago

9

Write a generic exponential growth function A(t) that has initial value P and an annual growth rate (expressed as a decimal) of

r.

1 answer:

Anvisha [2.4K]3 years ago

3 0

Answer:

A(t) = P(1+r)^t

Step-by-step explanation:

A(t): amount after

P: initial value

t: time

r: growth as a decimal

This example shows compounding the growth once per year

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HELP ME PLZ AND IF YOU HELP ME 10 POINTS

mafiozo [28]

Answer:

3(c+4)

Step-by-step explanation:

It would usually be 3 x 4, but since there is a variable, it can`t be multiplied (unless you know what the value of the variable is). Thus the expression is 3(c+4).

6 0

3 years ago

Read 2 more answers

Need urgent help in mathematics!! Will mark brainliest!!!!

Nataly [62]

₇P₅ is a permutation. you can find this setting in the statistics menu of your scientific calculator (i think).

If you don't have a scientific calculator, you can use the following formula:

nPr = $\frac{n!}{(n - r)!}$

₇P₅ = $\frac{7!}{(7 - 5)!}$

= $\frac{7}{2!}$

= $\frac{7 * 6 * 5 * 4 * 3 * 2!}{2!}$

= 7 * 6 * 5 * 4 * 3

= 2520

Answer: C

3 0

4 years ago

Read 2 more answers

Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

4 years ago

23 kg of potatoes cost $138. How much would 7 kg cost? lol i haven't done this in a while and there just now telling us to do it

Aleonysh [2.5K]

Answer:

$42

Step-by-step explanation:

138/23 = $6 per kg

6*7 = 42

8 0

3 years ago

Directions:Use the product rule to simplify the following monomials.

attashe74 [19]

13) Answer: $-3a^3b^2$

$\left(-6a^2b\right)\left(\frac{1}{2}ab\right)$

$=-6a^2b\frac{1}{2}ab$

$=-6a^2b^2\frac{1}{2}a$

$=-6a^3b^2\frac{1}{2}$

$=-3a^3b^2$

14) Answer: $8y^5x$

$12y\left(\frac{2}{3}xy^4\right)$

$=12\cdot \frac{2}{3}xy^{1+4}$

$=12\cdot \frac{2}{3}xy^5$

$=\frac{2\cdot \:12xy^5}{3}$

$=\frac{24y^5x}{3}$

$=8y^5x$

6 0

3 years ago