MATHEMATICS HIGH SCHOOL

20a + 158 + 6c = 20 (6) +15(4) + 6(8)
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Answers

Answer 1

Answer:

a=7/2-3/10

Step-by-step explanation:

20a +158+6c= 120+60+48

20a +158+6c=228

20a =228-158-6c

20a =70-6c

a= 7/2 - 3/10

Answer 2

Answer:

Answer:

120 + 60 + 48 would be the correct answer !

Step-by-step explanation:

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Caroline bought 20 shares of stock at 101/2, and after 10 months the value of the stocks was 111/4. If Caroline were to sell all her shares of this stock, how much profit would she make

Answers

Answer:

$15 profit would she make

Step-by-step explanation:

Given the statement: Caroline bought 20 shares of stock at $10(1)/(2)$ and after 10 months the value of the stock was $11(1)/(4)$ if Caroline were to sell all her shares of this stock.

$\text{Profit} = 11(1)/(4)-10(1)/(2) = 11.25- 10.5 = 0.75$

Number of share = 20.

$\text{Total profit } = 0.75 * 20 =\$15$

Therefore, the profit would she make is, $15

What two numbers multiply to 44 and add up to 12?

Answers

This pattern of question is always coming up. Since we can't easily guess, then let us set up simultaneous equation for the statements.

let the two numbers be x and y.

Multiply to 44.    x*y = 44 ..........(a)

Add up to 12. x + y = 12 .........(b)

From (b)

y = 12 - x .......(c)

Substitute (c) into (a)

x*y = 44

x*(12 - x) = 44

12x - x² = 44

-x² + 12x = 44

-x² + 12x - 44 = 0.

Multiply both sides by -1

-1(-x² + 12x - 44) = -1*0

x² - 12x + 44 = 0.

This does not look factorizable, so let us just use quadratic formula

comparing to ax² + bx + c = 0, x² - 12x + 44 = 0, a = 1, b = -12, c = 44

x = (-b + √(b² - 4ac)) /2a   or (-b - √(b² - 4ac)) /2a

x = (-(-12) + √((-12)² - 4*1*44) )/ (2*1)

x = (12 + √(144 - 176) )/ 2

x = (12 + √-32 )/ 2

√-32 = √(-1 *32) = √-1 * √32 = i * √(16 *2) = i*√16 *√2 = i*4*√2 = 4i√2

Where i is a complex number. Note the equation has two values. We shall include the second, that has negative sign before the square root.

x = (12 + √-32 )/ 2 or (12 - √-32 )/ 2

x = (12 + 4i√2 )/ 2 (12 - 4i√2 )/ 2

x = 12/2 + (4i√2)/2 12/2 - (4i√2)/2

x = 6 + 2i√2 or    6 - 2i√2

Recall equation (c):

y = 12 - x, When x = 6 + 2i√2, y = 12 - (6 + 2i√2) = 12 - 6 - 2i√2 = 6 - 2i√2

When x = 6 - 2i√2, y = 12 - (6 - 2i√2) = 12 - 6 + 2i√2 = 6 + 2i√2

x = 6 + 2i√2, y = 6 - 2i√2

x = 6 - 2i√2, y = 6 + 2i√2

Therefore the two numbers that multiply to 44 and add up to 12 are:

6 + 2i√2 and 6 - 2i√2

$xy=44\n x+y=12\n\n xy=44\n x=12-y\n\n (12-y)y=44\n 12y-y^2=44\n y^2-12y+44=0\n y^2-12y+36+8=0\n (y-6)^2=-8$

No solutions in real numbers.

In complex numbers:
$(y-6)^2=-8\n y-6=-√(-8) \vee y-6=√(-8)\n y=6-2\sqrt2 i \vee y=6+2\sqrt2 i\n\n x=12-(6-2\sqrt2i) \vee x=12-(6+2\sqrt2i)\n x=12-6+2\sqrt2i \vee x=12-6-2\sqrt2i\n x=6+2\sqrt2i \vee x=6-2\sqrt2i$

These numbers are $6-2\sqrt2i$ and $6+2\sqrt2i$ .

What is the greatest common factor of 8m, 36m3, and 12?

4
8
4m
8m

Answers

The greatest common factor is exactly as it sounds: the greatest factors of two or more expressions.

Factorize the numbers and identify all common factors. To get the GCF multiply all common factors:

$8m=2\cdot 2\cdot 2\cdot m,\n36m^3=2\cdot 2\cdot 3\cdot 3\cdot m\cdot m\cdot m,\n 12=2\cdot 2\cdot 3$ .

Common factors are: 2 and 2, their product is 2·2=4, then the greatest common factor is 4.

Hello,
the gcd is 4 .

At the beginning of the year, Noah had $90 in savings and saved an additional $10 each week thereafter. Jace started the year with $50 and saved $20 every week. Let NN represent the amount of money Noah has saved tt weeks after the beginning of the year and let JJ represent the amount of money Jace has saved tt weeks after the beginning of the year. Graph each function and determine the number of weeks after the beginning of the year until Noah and Jace have the same amount of money saved.

Answers

Answer:

Noah and Jace will have the same amount of money saved 4 weeks after the beginning of the year. This is the point of intersection on the graph of the two functions.

Step-by-step explanation:

To graph the functions that represent the amount of money Noah and Jace have saved, we can set up their savings functions as follows:

Noah's savings function (NN):

NN(tt) = $90 (initial savings) + $10 (weekly savings) * tt

Jace's savings function (JJ):

JJ(tt) = $50 (initial savings) + $20 (weekly savings) * tt

Now, let's graph these functions:

Noah's Savings Function (NN):

Initial savings (y-intercept): $90

Weekly savings rate (slope): $10

Jace's Savings Function (JJ):

Initial savings (y-intercept): $50

Weekly savings rate (slope): $20

To find the number of weeks after the beginning of the year until Noah and Jace have the same amount of money saved, we can set the two functions equal to each other and solve for "tt":

NN(tt) = JJ(tt)

$90 + $10 * tt = $50 + $20 * tt

Now, solve for "tt":

$90 - $50 = $20 * tt - $10 * tt

$40 = $10 * tt

tt = $40 / $10

tt = 4

In 36 years, amy will be 3 times as old as she is right now. how old is she now?