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2 years ago

Asiandoll

Mathematics

1 answer:

Sholpan [36]2 years ago

5 0

Given h(x) = 2x- 5 and t(x) = 6x + 4

PART(A)

h(x) + t(x) is simply the sum of two functions so let's add the expression given for hx and tx

h(x) + t(x) = 2x -5 + 6x + 4

= 8x -1 Answer

PART (B)

( h. t)(x) is simply the product of two functions :

(2x-5)( 6x+4) = 2x* 6x + 2x*4 -5* 6x -5*4

= 12x^2 + 8x -30x -20

= 12x^2 -22x -20 Answer

PART (C)

When it is h( tx) it is composite function where tx is the inner function in the function hx

we rewrite the function hx by replacing x in it with tx = 6x+4

h (x)= 2x - 5

on left side x is replaced with t(x) and on right side we replace x with 6x+4

h( tx)= 2( 6x +4) - 5

h(tx) = 12 x + 2*4 - 5

h( tx) + 12x + 8 - 5

h( tx) = 12x + 3

Answer 12x +3

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A culture of bacteria has an initial population of 9300 bacteria and doubles every 3

sp2606 [1]

Answer:

The approximate population of bacteria in the culture after 10 hours is 93,738.

Step-by-step explanation:

<h3>General Concepts:</h3>

Exponential Functions.
Exponential Growth.
Doubling Time Model.
Logarithmic Form.

BPEMDAS Order of Operations:

Brackets.
Parenthesis.
Exponents.
Multiplication.
Division.
Addition.
Subtraction.

<h2>Definitions:</h2>

We are given the following Exponential Growth Function (Doubling Time Model), $\displaystyle\mathsf{P_{(t)}\:=\:P_0\cdot2^{(t/d)}}$ where:

$\displaystyle\sf{P_t\:\:\rightarrow}$ The population of bacteria after “<em>t </em>” number of hours.
$\displaystyle\sf{P_0 \:\:\rightarrow}$ The initial population of bacteria.
$\displaystyle{t \:\:\rightarrow}$ Time unit (in hours).
$\displaystyle{\textit d \:\:\rightarrow}$ Doubling time, which represents the amount of time it takes for the population of bacteria to grow exponentially to become twice its initial quantity.

<h2>Solution:</h2>

<u>Step 1: Identify the given values.</u>

$\displaystyle\sf{P_0\:=}$ 9,300.
<em>t</em> = 10 hours.
<em>d</em> = 3.

<u>Step 2: Find value.</u>

1. Substitute the values into the given exponential function.

$\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}$

$\displaystyle\mathsf{\longrightarrow P_{(10)} = 9300\cdot2^{(10/3)}}$

2. Evaluate using the BPEMDAS order of operations.

$\displaystyle\mathsf{P_{(10)} = 9300\cdot2^{(10/3)}\quad \Longrightarrow BPEMDAS:\:(Parenthesis\:\:and\:\:Division).}$

$\displaystyle\sf P_{(10)} = 9300\cdot2^{(3.333333)}\quad\Longrightarrow BPEMDAS:\:(Exponent).}$

$\displaystyle\sf P_{(10)} = 9300\cdot(10.079368399)\quad \Longrightarrow BPEMDAS:(Multiplication).}$

$\boxed{\displaystyle\mathsf{P_{(10)} \approx 93,738.13\:\:\:or\:\:93,738}}$

Hence, the population of bacteria in the culture after 10 hours is approximately 93,738.

<h2>Double-check:</h2>

We can solve for the amount of <u>time</u> <u>(</u><em>t</em> ) it takes for the population of bacteria to increase to 93,738.

1. Identify given:

$\displaystyle\mathsf{P_{(t)} = 93,738 }$ .
$\displaystyle\mathsf{P_0 = 9,300}$ .
<em>d </em>= 3.

2. Substitute the values into the given exponential function.

$\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}$

$\displaystyle\mathsf{\longrightarrow 93,378 = 9,300\cdot2^{(t/3)}}$

3. Divide both sides by 9,300:

$\displaystyle\mathsf{\longrightarrow \frac{93,378}{9,300} = \frac{9,300\cdot2^{(t/3)}}{9,300}}$

$\displaystyle\mathsf{\longrightarrow 10.07936840 = 2^{(t/3)}}$

4. Transform the right-hand side of the equation into logarithmic form.

$\boxed{\displaystyle\mathsf{\underbrace{ x = a^y}_{Exponential\:Form} \longrightarrow \underbrace{y = log_a x}_{Logarithmic\:Form}}}$

$\displaystyle\mathsf{\longrightarrow 10.07936840 = \bigg[\:\frac{t}{3}\:\bigg]log(2)}$

5. Take the <em>log</em> of both sides of the equation (without rounding off any digits).

$\displaystyle\mathsf{log(10.07936840) = \bigg[\:\frac{t}{3}\:\bigg]log(2)}$

$\displaystyle\mathsf{\longrightarrow 1.003433319 = \bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996)}$

6. Divide both sides by (0.301029996).

$\displaystyle\mathsf{\frac{1.003433319}{0.301029996} = \frac{\bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996) }{0.301029996}}$

$\displaystyle\mathsf{\longrightarrow 3.3333333 = \frac{t}{3}}$

7. Multiply both sides of the equation by 3 to isolate "<em>t</em>."

$\displaystyle\mathsf{(3)\cdot(3.3333333) = \bigg[\:\frac{t}{3}\:\bigg]\cdot(3)}$

$\boxed{\displaystyle\mathsf{t\approx10}}$

Hence, it will take about 10 hours for the population of bacteria to increase to 93,378.

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<span>(12xy)^2

Expand:

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Find the area of the parallelogram with the given vertices. k(1, 1, 3), l(1, 3, 5), m(6, 9, 5), n(6, 7, 3).

nekit [7.7K]

The area of the parallelogram is $\sqrt{132}$ .

<h3>What is a parallelogram?</h3>

A parallelogram is a quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure. Also, the interior angles on the same side of the transversal are supplementary. Sum of all the interior angles equals 360 degrees.

Given that,

vertices k(1, 1, 3), l(1, 3, 5), m(6, 9, 5), and n(6, 7, 3)

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KL = (1, 1, 3) - (1, 3, 5) = (0, 2, 2)

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