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o-na [289]
3 years ago
7

3x+3y=-3 solve using the elimination

Mathematics
1 answer:
timofeeve [1]3 years ago
5 0

Simplifying

3x + 3y = -3

Solving

3x + 3y = -3

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3y' to each side of the equation.

3x + 3y + -3y = -3 + -3y

Combine like terms: 3y + -3y = 0

3x + 0 = -3 + -3y

3x = -3 + -3y

Divide each side by '3'.

x = -1 + -1y

Simplifying

x = -1 + -1y

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STatiana [176]

Answer:

The net worth does not change.

Step-by-step explanation:

Net worth is the sum of all of your assets minus the liabilities.

If you start off with 2,500 in your savings account, then you use the money to buy a car the net worth does not change. You started with a net worth of 2,500 because money in your savings account in considered an asset. You then use that money to buy a car which is an asset, and the car contributes to your net worth.

7 0
3 years ago
Find the unit rate<br><br> 8 hits in 22 games
Aleksandr [31]

Answer:

0.363636

Step-by-step explanation:

Hits per game is hits / games

= 8 / 22

= 0.363636363

8 0
3 years ago
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

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

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
3 years ago
A. Plot 13.58 on a number line on which the long
Lerok [7]

Answer:

Se attached for plots and here is position of the numbers:

a. Plot 13.58 on a number line on which the long

tick marks are whole numbers.

  • 13 < 13.58 < 14

b. Plot 0.193 on a number line on which the long

tick marks are tenths.

  • 0.1 < 0.193 < 0.2

c. Plot 26.9999 on a number line on which the

long tick marks are thousandths.

  • 26.999 < 26.9999 < 27.000

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77julia77 [94]

Answer:

The sis that said hello 5 times higher said hello 10 times

Step-by-step explanation:

The first sis said hay online 5 times

The second sis said hello 5 times higher

5 times higher meaning she said hello 10 times in total

5 0
3 years ago
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