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3 years ago

Which set

Mathematics

1 answer:

pogonyaev3 years ago

5 0

Answer:

The correct answer is: c. {-3, 0, 4, 11}

Step-by-step explanation:

A function consists of ordered pairs. In the ordered pairs, the first element is the input and second value is the output.

All the possible set of inputs for a function are called domain.

Given function is:

{(-3, 6), (0, 2), (4, 7), (11, 15)}

The domain set will consist of the first elements of all the ordered pairs.

The domain of this function is: {-3,0,4,11}

Hence,

The correct answer is: c. {-3, 0, 4, 11}

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Solve using<br> elimination <br> X- 2y = 3<br> 3x - y = 2

FrozenT [24]

<u>Answer:</u> x = 1/5 and y= -7/5

Step-by-step explanation: Let's solve your system by elimination.

Multiply the first equation by -3,and multiply the second equation by 1.

-3 (x - 2y = 3)

1 (3x - y = 2)

Becomes:

-3x + 6y = -9

3x -y = 2

Add these equations to eliminate x:

5y = -7

Then solve 5y = -7

5y/7 = -7/5

y = -7/5

Now that we‘ve found y, let’s plug it back in and solve for x.

Write down an original equation:

x - 2y = 3

<h3>Substitute -7/5 for y in x - 2y = 3:</h3><h3>x - 2 (-7/5) = 3</h3><h3>x + 14/5 = 3</h3><h3>x 14/5 + -14/5 = 3 + -14/5 (Add (-14)/5 to both sides)</h3><h3>x = 1/5</h3><h3><u>Answer: </u></h3><h3>x = 1/5 and y = -7/5</h3><h3 /><h3 /><h3 />

6 0

3 years ago

Sarah is carrying out a series of experiments which involve using mcreasing amounts of a chemical. In the

Marianna [84]

Sarah is carrying out a series of experiments which involve using increasing amounts of a chemical. In the first experiment she uses 6g of the chemical and in the second experiment she uses 7.8 g of the chemical

(i)Given that the amounts of the chemical used form an arithmetic progression find the total amount of chemical used in the first 30 experiments

(ii)Instead it is given that the amounts of the chemical used for a geometric progression. Sarah has a total of 1800 g of the chemical available. Show that the greatest number of experiments possible satisfies the inequality: $1.3^N \leq 91$ and use logarithms to calculate the value of N.

Answer:

(a)963 grams

(b)N=17

Step-by-step explanation:

(a)

In the first experiment, Sarah uses 6g of the chemical

In the second experiment, Sarah uses 7.8g of the chemical

If this forms an arithmetic progression:

First term, a =6g

Common difference. d= 7.8 -6 =1.8 g

Therefore:

Total Amount of chemical used in the first 30 experiments

$S_n=\dfrac{n}{2}[2a+(n-1)d] \\S_{30}=\dfrac{30}{2}[2*6+(30-1)1.8] \\=15[12+29*1.8]\\=15[12+52.2]\\=15*64.2\\=963$ grams$

Sarah uses 963 grams in the first 30 experiments.

(b) If the increase is geometric

First Term, a=6g

Common ratio, r =7.8/6 =1.3

Sarah has a total of 1800 g

Therefore:

Sum of a geometric sequence

$S_n=\dfrac{a(r^N-1)}{r-1} \\1800=\dfrac{6(1.3^N-1)}{1.3-1} \\1800=\dfrac{6(1.3^N-1)}{0.3}\\$Cross multiply\\1800*0.3=6(1.3^N-1)\\6(1.3^N-1)=540\\1.3^N-1=540\div 6\\1.3^N-1=90\\1.3^N=90+1\\1.3^N=91$