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disa [49]
1 year ago
10

Which of the following is the domainof the given relation?{(-6, 3), (-4 ,5), (0, 0)}A {0, 4, -6}B {0, 3, 5)C {3, 4, 0}D {-6, -4,

5)
Mathematics
1 answer:
Bas_tet [7]1 year ago
6 0

The domain is the input, which is the x value in the relation (x, y)

Hence the domain in the given relation {(-6, 3), (-4 ,5), (0, 0) is :

{-6, -4, 0}

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The Sugar Sweet Company is going to transport its sugar to market. It will cost $3375 to rent trucks, and it will cost an additi
Ksju [112]

Answer:

C = 125S + $3750

Step-by-step explanation:

$ 3750  = truck rental;   $125 per ton of sugar transported.  C is the cost;  S is the number of tons transported. The equation relating C to S would be a linear equation like y = mx + b. This equation would be graphed in the first quadrant only. You would start with your y-intercept at (0, 3750). As x increases by 1, your y increases by 125 yielding these points:

(1, 3875)  (2, 4000)  (3, 4125)  etc.

This shows that for each increase by one ton of sugar, the cost goes up $125

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A company makes two mixtures of nuts: Mixture A and Mixture B. Mixture A contains 30% peanuts, 30% almonds and 40% cashews and s
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Emily has 32 grams of a 25% sugar syrup, but this is just too sweet for her! How much pure water should she add to have a 10% sy
GarryVolchara [31]
32 grams = water mass + sugar mass . 100% syrup=75% water+ 25%sugar 
original water content= 32*75/100=24 grams
original sugar content= 32*25/100= 8grams 

total mass at 10% syrup:
8grams/total mass=10%=0.1
total mass=8/0.1=80grams
total mass of water=80grams- 8grams of sugar content= 72 grams
total mass of water added=72grams- 24grams original water content= 48grams.

answer: 48grams

3 0
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5 0
2 years ago
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What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
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