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2 years ago

Let f(x) = 3x2 x − 3 and g(x) = x2 − 5x 1. find f(x) − g(x). select one:

Mathematics

1 answer:

serg [7]2 years ago

5 0

I hope this helps you

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The cubit is an ancient unit. Its length equals six palms. (A palm varies from 2.5 to 3.5 inches depending on the individual.) W

olasank [31]

Assuminhg the ark has a shoe-box (cuboid) shape it´s volume would be:

$V_{cuboid}=lenght*width*height$

We have this three measurements ( $l=300cubits;w=50cubits;h=30cubits)$ ) and it can be as simple as replacing them in the equation and solving but they are in all in $cubits$ . We will convert them to $ft$ because the problem requires the answer in $ft^{3}$ . In order to do this we will use the given equivalences:

$1cubit=6palms\\1palm=3.10in$

and another one:

$1ft=12in$

First we will convert from cubits to palms:

$l=300cubits*\frac{6palms}{1cubit}=1800palms\\w=50cubits*\frac{6palms}{1cubit}=300palms\\h=30cubits*\frac{6palms}{1cubit}=180palms\\$

now from $palms$ to $in$ :

$l=1800palms*\frac{3.10in}{1palm}=5580in\\w=300palms*\frac{3.10in}{1palm}=930in\\h=180palms*\frac{3.10in}{1palm}=558in\\$

now from $in$ to $ft$ :

$l=5580in*\frac{1ft}{12in}=465ft\\w=930in*\frac{1ft}{12in}=77.5ft\\h=558in*\frac{1ft}{12in}=46.5ft$

We can calculate the Volume now like this:

$V_{ark}=465ft*77.5ft*46.5ft=1675743.75ft^{3}$

The volume of trhe ark would be $1675743.75ft^{3}$

6 0

2 years ago

A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 3 feet, 55 feet awa

Juliette [100K]

Answer:

Step-by-step explanation:

55+14

7 0

2 years ago

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Sarah kicked a ball in the air. The function

Aleksandr-060686 [28]

Answer: The ball hits the ground at 5 s

Step-by-step explanation:

The question seems incomplete and there is not enough data. However, we can work with the following function to understand this problem:

$f=30 t- 6t^{2}$ (1)

Where $f$ models the height of the ball in meters and $t$ the time.

Now, let's find the time $t$ when the ball Sara kicked hits the ground (this is when $f=0 m$ ):

$0=30 t- 6t^{2}$ (2)

Rearranging the equation:

$6t^{2}-30 t=0$ (3)

Dividing both sides of the equation by $6$ :

$t^{2}-5 t=0$ (4)

This quadratic equation can be written in the form $at^{2}+bt+c=0$ , and can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=1$

$b=-5$

$c=0$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(0)}}{2(1)}$ (6)

Solving we have the following result:

$t=5 s$ This means the ball hit the ground 5 seconds after it was kicked by Sara.

6 0

2 years ago

One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number o

Zielflug [23.3K]

To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

8 0

2 years ago

I know twice as many jokes as my friend does. Together we know 60 different jokes. My friend and I both know the same three joke

Murljashka [212]

Answer:

You know 40 jokes and your friend knows 20 jokes

Step-by-step explanation:

Given you know twice as many jokes as your friend knows

Let the number of jokes your friend knows be x , then the number of jokes known by yourself will be 2x

Given together you both know 60 jokes

Total number of jokes = number of jokes known by yourself + number of jokes known by your friend

60 = x + 2x

x = 20

Therefore you know 40 jokes and your friend knows 20 jokes

5 0

2 years ago

Read 2 more answers