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3 years ago

What are the constants in this expression? -10.6+9/10+2/5m-2.4n+3m

Mathematics

1 answer:

UNO [17]3 years ago

7 0

Answer:

-10.6 and ⁹/₁₀

Step-by-step explanation:

The constants are the numbers that stand by themselves: -10.6 and ⁹/₁₀.

The <em>variables</em> are m and n.

The <em>coefficients</em> are the numbers in front of the variables: ⅖, 2.4, and 3.

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Gillian can spend up to $300 buying both cotton shirts and wool shirts for her club. The cotton shirts cost $15 each and the woo

Vaselesa [24]

Answer: Option 'A' is correct

Step-by-step explanation:

Since we have given that

Total money Gillian spend on both cotton shirts and wool shirts = $300

Cost of per cotton shirt = $15

Cost of per wool shirt = $ 20

According to question,

The number of wool shirts Gillian will buy will be at least 3, but less than half the number of cotton shirts she buys.

So, by option when we use the data of Option 'A'

Number of cotton shirts =13

Cost incurred on cotton shirt is given by

$13\times 15=195$

Number of wool shirts = 4

Cost incurred on wool shirt is given by

$4\times 20=80$

Total cost incurred on both the shirts is given by

$195+80=\$275$

If we use option 'B' we spend only $230

If we use option 'C' , it does not satisfy the question as it is mentioned in the question that number of wool shirt must be less than half of number of cotton shirt

If we use option 'D' , we spend $325 which is much higher than $300

Hence , option "A" is correct.

3 0

3 years ago

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I NEED THIS ASAP WHOEVER GETS FIRST GET BRAINLIEST

Rina8888 [55]

Answer:

It is A

Step-by-step explanation:

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3 years ago

Suppose that 55% of all adults regularly consume coffee,

ra1l [238]

A. To answer this question, directly multiply the probability of those who consume coffee regularly and those who consume carbonated soda.
P (coffee and carbonated soda) = 0.55 x 0.45
= 0.2475

b. Since there are 705 of them who consume at least one of these two products then, the probability of the randomly selected individual who does not consume at least one of them is equal to 1 - 0.70 = 0.30.

4 0

3 years ago

How many solutions do the equations y=3x+4 and 3y-6x=2 have?

Alekssandra [29.7K]

Answer:

1 solution

Step-by-step explanation:

You can solve these equations by using substitution. y = 3x+4, so 3y-6x=2 can also be written as 3(3x+4)-6x = 2 if you plug in the y value in the second equation. Then, simplify and you will get your x answer. If you plug that into the first equation, you will get your y answer.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Crm%5Cint_%7B0%7D%5E%7B%20%5Cinfty%20%7D%20%5Cleft%28%20%20%5Cfrac%7B%2

mylen [45]

Recall the geometric sum,

$\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}$

It follows that

$1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}$

So, we can rewrite the integral as

$\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Split up the integral at x = 1, and consider the latter integral,

$\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Substitute $x\to\frac1x$ to get

$\displaystyle \int_0^1 \frac{\frac1{x^2} + 1}{\frac1{x^4} + \frac1{x^2} + 1} \frac{\ln\left(1 + \frac1{x^{2021}}\right) - \ln\left(1 + \frac1x\right)}{\ln\left(\frac1x\right)} \, \frac{dx}{x^2}$

Rewrite the logarithms to expand the integral as

$\displaystyle - \int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2021}+1) - \ln(x^{2021}) - \ln(x+1) + \ln(x)}{\ln(x)} \, dx$

Grouping together terms in the numerator, we can write

$\displaystyle -\int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2020}+1)-\ln(x+1)}{\ln(x)} \, dx + 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

Substituting $x\to\frac1x$ again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 1010 \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 505 \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx$

We can neatly handle the remaining integral with complex residues. Consider the contour integral

$\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz$

where γ is a semicircle with radius R centered at the origin, such that Im(z) ≥ 0, and the diameter corresponds to the interval [-R, R]. It's easy to show the integral over the semicircular arc vanishes as R → ∞. By the residue theorem,

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4}\, dx = 2\pi i \sum_\zeta \mathrm{Res}\left(\frac{1+z^2}{1+z^2+z^4}, z=\zeta\right)$

where $\zeta$ denotes the roots of $1+z^2+z^4$ that lie in the interior of γ; these are $\zeta=\pm\frac12+\frac{i\sqrt3}2$ . Compute the residues there, and we find

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx = \frac{2\pi}{\sqrt3}$

and so the original integral's value is

$505 \times \dfrac{2\pi}{\sqrt3} = \boxed{\dfrac{1010\pi}{\sqrt3}}$

8 0

2 years ago