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Rainbow [258]
3 years ago
13

Fill in the Blank 1. 272 > 14 ​

Mathematics
1 answer:
LenaWriter [7]3 years ago
8 0

Answer:

19.4285714286 Step-by-step explanation: 272 divided by 14 can be calculated as follows. 272/14 math problems division

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without building the graph, find the coordinates of the point of intersection of the lines given by the equation y=3x-1 and 3x+y
DaniilM [7]
<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

  • y = 3x - 1
  • 3x + y = -7

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

\implies 3x + y = -7

\implies3x + (3x - 1) = -7

\implies3x + 3x - 1 = -7

Combine like terms as needed;

\implies 3x + 3x - 1 = -7

\implies 6x - 1 = -7

Add 1 to both sides of the equation;

\implies 6x - 1 + 1 = -7 + 1

\implies 6x = -6

Divide 6 to both sides of the equation;

\implies \dfrac{6x}{6}  = \dfrac{-6}{6}

\implies x = -1

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

\implies y = 3(-1) - 1

\implies     \ \ = -3 - 1

\implies     = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.

\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.

\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

\implies 0 = (6x) + (6)

\implies0 = 6x + 6

This problem is now an algebraic problem. Isolate "x" to determine its value.

\implies 0 - 6 = 6x + 6 - 6

\implies -6 = 6x

\implies -1 = x

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

\implies y = 3(-1) - 1

\implies y = -3 - 1

\implies y = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

5 0
2 years ago
Fred had 108!dollars to spend on 6 books after buying them he had 18 dollars how much did each book cost?
lana [24]
Each book costs $15. since he had 18 dollars left, you can subtract 18 from 108. then divide that answer by 6. 90/6 = 15.
Hope that helped!
7 0
3 years ago
(I NEED IT NOWWWWW HURRYYYY)You have half of your pens in your backpack. You give half of these to your friend, Alex. What fract
Margarita [4]

Answer:

1/4 of ur pens

Step-by-step explanation:

8 0
3 years ago
PLEASE HELP
LekaFEV [45]

Answer: one solution

Step-by-step explanation:

CONCEPT:

- One solution is when the final variable would be able to find a solution.

- Infinite solution is when the equations result in 0=0, meaning any number will fit

- No solution is when the equation results in both sides on equal.

SOLVE:

Given

3/4x-7=3(1/6x+1)

Expand Parenthesis

3/4x-7=1/2+3

Multiply both sides by the LCM of fractions (Least Common Multiple)

4(3/4x-7)=4(1/2x+3)

3x-28=2x+12

Subtract both sides by 2x

3x-28-2x=2x+12-2x

x-28=12

Add both sides by 28

x-28+28=12+28

x=40

Since we are able to get exactly one solution, there is only one solution.

Hope this helps!! :)

Please let me know if you have any quesionts

5 0
3 years ago
The solutions to a function are what points on the coordinate plane when you graph the line?​
vodka [1.7K]

The solutions to a function are the points where the graph of the function crosses the x-axis, but only if the solutions are real numbers. If the function has imaginary solutions, then you don't see those on the graph.

4 0
3 years ago
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