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3 years ago

10

The ice cream shop has 80 different possible sundaes consisting of 1 flavor of ice cream, 1 syrup, and 1 candy topping. If the i

ce cream parlor has 5 different syrups and 4 different candy toppings, then how many flavors of ice cream does it have?

1 answer:

adoni [48]3 years ago

4 0

<span>4: 80*5*4 </span>
<span>= 1600 diff ice creams flavors. yum.

</span>

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HELP PLEASE<br> Use elimination to solve the system of equations<br> 2x + 3y = 9 <br> x + 5y = 8

ikadub [295]

Answer:

y=5/7 x=31/7

Step-by-step explanation:

Make the x values the opposite of each other:

-2(x+5y=8)

-2x-10y=-16

2x+3y=9

Combine like terms and solve for y:

-7y=-5

y=5/7

Plug y in to find x.

x+5(5/7)=8

x+25/7=8

56/7-25/7=x

x=31/7

5 0

2 years ago

Joyce is trying to solve the equation y = x2 − 8x + 7 using the quadratic formula. She has made an error in one of the steps bel

kobusy [5.1K]

Answer:

x = 1 , 7

Step-by-step explanation:

Solution:-

- The given equation is as follows:

y = x^2 - 8x + 7

- We can solve the above equation by either making factors or by using Quadratic formula.

Factor Approach:

- Using the constant "7" at the end of the quadratic equation we will determine two integer multiples such that their additions/subtraction results in "-8".

- So the only factor of "7" are:

7 x 1 = 7

-7 x -1 = 7

- We see that addition/subtraction of first (7 , 1 ) does not results in "-8", However, the sum of ( -1 , -7 ) = -1 - 7 = -8. So the correct factors are ( -1 , -7 ). So we replace "-8x" with our factors "-1x" and "-7x":

x^2 -x -7x + 7 = 0

- Take common multiples out of pair of two terms:

x*(x-1) -7*(x-1) = 0

(x-7)*(x-1) = 0

- So we equate each term in bracket with "0" and evaluate the values of x:

(x-7) = 0 , x = 7

(x-1) = 0 , x = 1

- So the solution to the quadratic equation is:

x = 1 , 7

4 0

3 years ago

Can u answer these for me with the work shown

babymother [125]

Answer:

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}$

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}$

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}$

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}$

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1$

$\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}$

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x$

Step-by-step explanation:

Required

Simplify

Solving (1):

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}$

Factorize the numerator and the denominator

$\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}$

Factor out x+2 at the numerator

$\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}$

Express x^2 - 9 as difference of two squares

$\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}$

$\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}$

Expand the denominator

$\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}$

Factorize

$\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}$

$\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}$

Cancel out same factors

$\frac{(x+3)}{x}$

Hence:

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}$

Solving (2):

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2}$

Expand the numerator and factorize the denominator

$\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}$

Factorize the numerator

$\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}$

Factor out x - 2

$\frac{(3x + 1)(x - 2)}{x^2(x- 2)}$

Cancel out x - 2

$\frac{3x + 1}{x^2}$

Hence:

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}$

Solving (3):

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}$

Express x^2 - 9 as difference of two squares

$\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}$

Factorize all:

$\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}$

Cancel out x + 3 and 3 + x

$\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}$

$\frac{3 - x}{x- 3} * \frac{5}{2x}$

Express $3 - x$ as $-(x - 3)$

$\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\$

$-1 * \frac{5}{2x}$

$-\frac{5}{2x}$

Hence:

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}$

Solving (4):

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}$

Expand x^2 - 6x + 9 and factorize 5x - 15

$\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}$

Factorize

$\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}$

$\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}$

Cancel out x - 3

$\frac{(x -3)}{5} / \frac{5}{3-x}$

Change / to *

$\frac{(x -3)}{5} * \frac{3-x}{5}$

Express $3 - x$ as $-(x - 3)$

$\frac{(x -3)}{5} * \frac{-(x-3)}{5}$

$\frac{-(x-3)(x -3)}{5*5}$

$\frac{-(x-3)^2}{25}$

Hence:

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}$

Solving (5):

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}$

Factorize the numerator and expand the denominator

$\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}$

Factor out x - 1 at the numerator and factorize the denominator

$\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}$

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

$\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}$

$x +1$

Hence:

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1$

Solving (6):

$\frac{9x^2 + 3x}{6x^2}$

Factorize:

$\frac{3x(3x + 1)}{3x(2x)}$

Divide by 3x

$\frac{3x + 1}{2x}$

Hence:

$\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}$

Solving (7):

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}$

Change / to *

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}$

Expand

$\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}$

Factorize

$\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}$

$\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}$

Cancel out x - 2 and x - 1

$\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}$

Cancel out x

$\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}$

$\frac{12x^2}{4x}$

$3x$

Hence:

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x$

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2 years ago

Find the following expression by multiplying.<br><br> (a + b)^2

Hatshy [7]

You multiply it together like (a+b)(a+b)
the answer would be a^2 + 2ab + b^2

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3 years ago

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1: x - (-3) is the same as x + (-3).

lyudmila [28]

1. False 2.True 3.20 4.13 5.-23 6. True 7. should be false because domain is a range of values and y- value is just one. 8.True

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2 years ago

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