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notsponge [240]
3 years ago
6

Please use method of substitution numbers 9;10;11;12

Mathematics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

1.    x = 1, y = 2.

12.  x = 1, y = -3.

Step-by-step explanation:

I won't do all these for you, just the first and the last. The method is similar  for all 12.

1.

y = x + 1

x + y = 3

Substitute y = x + 1 into the second equation:

x + y = 3 so:

x + (x + 1) = 3

2x + 1 = 3

2x + 1 -1 = 3 - 1

2x = 2

x = 1.

Now substitute x = 1 into the first equation to get the value of y:

y = x + 1

y = 1 + 1

y = 2.

12.

4x = y + 7

3x + 4y + 9 = 0

Solve for y in the first equation:

4x = y + 7                Subtract 7 from both sides of the equation:

4x - 7 = y + 7 - 7

y = 4x - 7.

Now substitute  y = 4x - 7 in the second equation:

3x + 4y + 9 = 0   so:

3x + 4(4x - 7) + 9 = 0

3x + 16x - 28 + 9 = 0

19x - 19 =  0

19x - 19 + 19 = 0 + 19

19x = 19

x = 19/19 = 1.

Now substitute x = 1 in the first equation:

4x = y + 7

4*1 = y + 7              Subtract 7 from both sides of the equation:

4 - 7 = y + 7 - 7      

-3 = y.

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Step-by-step explanation:

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Step-by-step explanation:

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Find the area of a rectangle with a perimeter of 12 meters and a base of 4 meters
stepladder [879]
Considering the perimeter (P)
p = 2 \times b + 2 \times h
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3 years ago
Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind
Gemiola [76]

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

7 0
3 years ago
Derrick will need $39,500 in 10 years for college tuition. How much should his parents invest now at 9.5% annual interest, compo
erastova [34]

Answer:

His parents should invest $15,278.16 to reach this goal ⇒ 4th answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Derrick will need $39,500 in 10 years for college tuition

∴ The future amount is $39,500

∴ The time for investment is 10 years

- P is the money his parents invest now at 9.5% annual interest,

 compounded daily

∴ The rate is 9.5% per year compounded daily

- The formula of the compounded interest is:

  A=P(1+\frac{r}{n})^{nt} , where

# A is the future value of money

# P is the value of investment

# r is the rate of interest in decimal

# t is the time of investment

# n is the period of the time

∵ A = $39,500

∵ t = 10

∵ r = 9.5/100 = 0.095

∵ n = 365 ⇒ compounded daily

- Lets use the formula above to find P

∴ 39500=P(1+\frac{0.095}{365})^{365*10}

∴ 39500=p(2.58539)

- Divide both sides by 2.58539

∴ P = $15278.16

∴ His parents should invest $15,278.16 to reach this goal

7 0
3 years ago
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