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2 years ago

Evaluate the expression 6(x) when x equals 3. Pls... I need help.

Mathematics

1 answer:

Leto [7]2 years ago

7 0

Answer:

Step-by-step explanation:

substitute x = 3 into the expression

6x = 6(3) = 6 × 3 = 18

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Determine whether each relationship represented by the

9966 [12]

Set number 1, 2, 3, and 4 or (a,b,c,d) is a function
EXPLANATION:
in a function you can’t have two of the same x values.
hope this helped :)

8 0

3 years ago

Find the amount in a continuously compounded account for the following condition. Principal, $4000; Annual interest rate, 5.

soldi70 [24.7K]

<h3>Answer: 4745.96 dollars</h3>

=======================================================

Explanation:

We have this given info

P = 4000 = principal
r = 0.057 = annual interest rate in decimal form
t = 3 = number of years

Use this to plug into the formula below

$A = Pe^{r*t}\\\\A = 4000*e^{0.057*3}\\\\A \approx 4,745.96299608713\\\\A \approx 4,745.96\\\\$

You'll need your calculator, and the calculator needs the "e" button.

The "e" refers to the special constant 2.718... which is similar to pi = 3.14...

7 0

2 years ago

10 ft А - “不 6 ft v 10 ft - 12 ft B C | 12 ft 4 ft 4 ft

Phoenix [80]

Answer:

Step-by-step explanation:

case I added it and I m right

8 0

2 years ago

The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and

brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

$y=0.00035x^{2}$

Compute the first order derivative of y as follows:

$y=0.00035x^{2}$

$\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]$

$=2\cdot 0.00035x\\\\=0.0007x$

Now, it is provided that |x | ≤ 605.

⇒ -605 ≤ x ≤ 605

Compute the arc length as follows:

$\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx$

$=\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\$

Now, let

$x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u$

$\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u$

$={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}$

Plug in the solved integrals in Arc Length and solve as follows:

$\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\$

$=1245.253707795227\\\\\approx 1245.25$

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

7 0

3 years ago

Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all

Oduvanchick [21]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

Given values:

$38, 40, 41, 45, 48, 48, 50, 50, 51, 51, 52, 52, 53, \\\\$ $54, 55, 55, 55, 56, 56, 57, 59, 59, 59, 62, 62, 62, \\ 63, 64, 65,66, 66, 67, 67, 69, 69, 71, 77, 78, 79, 79$

Total staff=40

In point a:

To calculate the median number first we arrange the value into ascending order and then collect the even numbers of calls that were also made. Its average of the middle terms is thus the median.

The midterms =55 and 59 so,

Median = $\frac{55+59}{2} = \frac{114}{2}=57$

In point b:

First-quarter $Q_1 = \frac{1}{4} \text{number of 8th call}$

$=\frac{1}{4} \times 30 th \\\\= 7.5 th$

The first quarterlies are $7.5th \ \ that \ is = (7+0.5)th\ \ term$

therefore the multiply of 0.5 by calculating the difference of the 7th and 8th term are:

$=0.5 \times 0= 0 \\\\\to Q_1 = 50+0=50 \\\\\to Q_3=\frac{3}{4} \text{number of 8th call} \\\\=\frac{3}{4} \times 30 th \\\\=22.5 \ th = (22+0.5)th \ \ term$

therefore the it is multiply by the 0.5 for the difference of the 22nd and 23rd term:

$= 0.5 \times 1=0.5 \\\\\to Q_3= 63+0.5=63.5$

In point c:

First decile $D_1 = \frac{1}{18} \text{number of 8th call}$

$= \frac{1}{10} \times 30 th \\\\= 3rd \ \ term\\\\\to D_1= 41 \\\\\to D_9= \frac{9}{10} \times 8\ th\ call\\\\=\frac{9}{10} \times 30 th \\\\=27th \ \ term\\\\\to D9 = 67$

In point d:

quartiles are:

$Q_1= 51.25 \\\\Q_3=66$

The right answers for the decile are:

$D_1=45.3 \\\\ D_9=76.4$

As for $D) P_{33} = 53.53$ will be available.

7 0

2 years ago