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2 years ago

Which expression is equivalent to 6f+f?

Mathematics

2 answers:

AleksAgata [21]2 years ago

6 0

Answer:

$f + 5f + f$ is equivalent to $6f + f$ .

Step-by-step explanation:

You can plug in any numbers to both expressions and they will always have the same outcome.

For example, let's plug in 2.

$6f + f\\6(2) + 2\\12 + 2 = 14$

$f + 5f + f\\2 + 5(2) + 2\\2+10+2 = 14$

Hope this helps! :)

Elena L [17]2 years ago

3 0

Answer:

Step-by-step explanation:

This is because 6f+f is just 7f.

f f f f f f+f = fffffff

Soooo, ONLY A gives us 7f when added together again.

f+5f+f=7f.

f+ f f f f f +f = fffffff

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A random sample is to be selected from a population that has a proportion of successes p=.60. When n=400, what is the probabilit

Nata [24]

Answer:

0.45134

Step-by-step explanation:

Given that :

p = 0.6

n = 400

Probability that sample. Proportion falls between 0.59 and 0.62

Using Normal approximation :

Mean (m) = n * p = 400 * 0.6 = 240

Standard deviation (s) = sqrt(pq/n)

q = 1 - p = 1 - 0.6 = 0.4

s = sqrt((0.6 * 0.4) / 400) = 0.0244948

P(0.59 < p < 0.62) :

(x - m) / s

P((0.59 - 0.6) / 0.0244948) < p < P((0.62 - 0.6) / 0.0244948)

P(Z < −0.408249) < p < P(Z < 0.8164998)

Using the Z probability calculator :

0.79289 - 0.34155 = 0.45134

3 0

3 years ago

An aerospace company has submitted bids on two separate federal government defense contracts. The company president believes tha

gulaghasi [49]

Answer:

1)28.8%

2)28.8%

3)43.6%

Step-by-step explanation:

1))Probability of wining both = Probability that first contract is won* Probability that second contract is won = 0.4 * 0.72 = 0.288 = 28.8%

2)Probability of losing = 1 - Probability of winning

Probability that they will lose both contracts = Probability that first contract is lost * Probability that second contract is lost = (1 - 40/100) * (1 - 54/100) = 0.276 = 27.6%

3)Probability that any one is won = 1 - both contracts are won - both contracts are lost = 1 - 0.276 - 0.288 = 0.436 = 43.6%

7 0

3 years ago

Each score in a set of data is multiplied by 3, and then 4 is subtracted from the result. If the original mean is 10 and the ori

olganol [36]

The new mean and standard deviation is 26 and 15, when each score in data set is multiplied by 5 and then 7 is added.

According to the question,

Original mean is 10 and original standard deviation is 5 . In order to find to new mean and standard deviation when each score in data set is multiplied by 5 and then 7 is added.

First "change of scale" when every score in a data set is multiplied by a constant, its mean and standard deviation is multiplied by a same constant.

Mean: 10*3 = 30

Standard deviation: 5*3 = 15

Secondly "change of origin" when every score in a data set by a constant, its mean get added or subtracted by the same constant and standard deviation remains constant.

Applying change of origin in the above mean and standard deviation

Mean: 30 - 4 = 26

Standard deviation: Remains same = 15

Hence, the new mean and standard deviation is 26 and 15, when each score in data set is multiplied by 5 and then 7 is added.

Learn more about Mean and standard deviation here

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5 0

2 years ago

How can i solve this equation to find x?

34kurt

Answer:

x = 25

Step-by-step explanation:

We know that A and M are on a line, which is 180 degrees, and the little square means 90 degrees. So, ∠RAM = 180 - 90 = 90 degrees.

∠RAM = ∠RAX + ∠XAM

90 = (2x - 10) + (-3x + 125)

Now, simply combine like terms and solve for x:

90 = 2x - 3x - 10 + 125

90 = -x + 115

x = 115 - 90 = 25

Thus, x = 25.

<em>~ an aesthetics lover</em>

5 0

4 years ago

Read 2 more answers

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago