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jarptica [38.1K]
3 years ago
15

How many solutions does the following equation have?

Mathematics
2 answers:
Katen [24]3 years ago
5 0
No solutions

-5(2+1)=-22+10
-10-5=-12
-15=-12
Alik [6]3 years ago
3 0

Answer:

no solutions

Step-by-step explanation:

-5 (2+1) = -22 +10

-15 = -12

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Tony’s club is selling oranges to raise money. For every box they sell, they get 1 1/8 dollars profit. They have sold 75 boxes a
dmitriy555 [2]

Tony’s club is selling oranges to raise money. For every box they sell, they get 1 1/8 dollars profit. They have sold 75 boxes already. How many more boxes must they sell to raise 180 dollars?

Answer: We are given:

The amount of money Tony's club get for every box they sell =1 \frac{1}{8} =1.125 dollar

They amount of money Tony's club has raised by selling 75 boxes is:

75 \times 1.125=84.375 dollars

The amount of money Tony's club is required to raise = 180 dollars

The remaining need to be raised is :

180-84.375=95.625 dollars

Therefore, the number of more boxes to be sold are:

\frac{95.625}{1.125}=85

Hence, 85 more boxes they must sell to raise 180 dollars

8 0
3 years ago
Mika wants to buy two types of cookies. The vanilla cookies cost $4 per box and the peanut butter cookies cost $6 per box. She w
ollegr [7]

Answer:

A & B

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
The line models Maria's saving account. She started with 7 in her account. She saves three each day. Write an equation for the l
artcher [175]

Answer:

y=7+3x

Step-by-step explanation:

Since she already has 7 in her account and she ADDS 3 more each day you would multiply 3 by x which represents the number of days and add that to seven because she already had it. This would equal to Y which represents the total in her account

8 0
3 years ago
① 5J+2S =12.20 ② 5J+ 10S =15.80 elimination method
hram777 [196]

Answer:

<h2>S = 9/20</h2><h2>J = 113/50</h2>

Step-by-step explanation:

5J +2S =12.20 ........(1)\\5J+10S=15.80......(2)\\Multiply \:equation \:(1)\:by\:the\:coefficient\:of\:x\:in equation\:(2)\\\\Multiply\:equation(2)\: by\: the\: coefficient\:of \:x\: in \:equation (1)\\\\5J +2S =12.20 ........(1) \times 5\\5J+10S=15.80......(2)\times 5\\\\25J+10S=61......(3)\\25J+50S = 79......(4)\\Subtract\:equation\:(4)\:from\: equation\: (3)\\-40S =-18\\\frac{-40S}{-40}=\frac{-18}{-40}\\  S = 9/20\\

Substitute \:9/20\:for \:x \:in\:equation\:(1)\:or \:equation\:(2)\\5J+2S = 12.20\\5J +2(9/20) =12.20\\5J +9/10=12.20\\5J=12.20-9/10\\5J=113/10\\Cross-Multiply\\50J = 113\\J = 113/50

4 0
3 years ago
PLEASE ANSWER QUICKLY
Olegator [25]

Answer:

x=3

Step-by-step explanation:

Set equations equal to each other:

x^3-3x^2+2=x^2-6x+11

Subtract two from both sides

x^3-3x^2=x^2-6x+9
Factor out x^2 from the left side

x^2(x-3)=x^2-6x+9

Factor the right side, look for two factors of 9, which add to -6, which is just -3 and-3

x^2(x-3) = (x-3)(x-3)

So as you can see here, we have both sides in factored form, and both have a factor of (x-3) on both sides, and if one of the factors is zero, then the entire thing becomes zero, since 0 * anything= zero.

this means a solution would be at x=3, since it makes the right side 0 * 0 = 0, and the left side (3)^2 * 0 = 0

It's important to notice this, since had we divided by x-3 to make the equation a quadratic, we would've excluded this real solution, because when dividing by x-3 the solution x=3 makes it 0, so we would've been dividing by zero.

So one of the real solutions is at x=3

6 0
2 years ago
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