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Lana71 [14]
3 years ago
15

PLEASE HELP VERY URGENT

Mathematics
1 answer:
gavmur [86]3 years ago
6 0

Answer:

B) (2, 1)

Step-by-step explanation:

−x + 3 = 2x - 3

-2x - 2x

___________

−3x + 3 = −3

- 3 - 3

_________

−6 = −3x

__ ___

−3 −3

2 = x [Plug this back into both equations above to get the y-coordinate of 1]; 1 = y

I am joyous to assist you anytime.

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Andrew claims the initial value and y - intercept are the same thing on a graph. Is he correct?
Agata [3.3K]

Answer:

We conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

Step-by-step explanation:

We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.

In other words, the initial value is the y-value or output of the point at x = 0

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the initial value of the equation y = 2x+1 is: y = 1

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Thus, at y = 1, the line meets the y-axis.

Hence, the initial value of the line is: y = 1

Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the y-intercept of y = 2x+1 is y = 1.

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Therefore, the y-intercept of y = 2x+1 is y = 1.

Conclusion:

Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

6 0
3 years ago
9 + 3x - 8<br> can you simplify ?
n200080 [17]

Answer:

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Step-by-step explanation:

9+3x-8

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Musya8 [376]

Answer:

C.  647 square units

Step-by-step explanation:

To find the shaded area, subtract the area of the unshaded square from the area of the octagon.

<u>Area of the octagon</u>

\textsf{Area of a regular polygon}=\dfrac{n\:l\:a}{2}

where:

  • n = number of sides
  • l = length of one side
  • a = apothem

Given:

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  • l = 13
  • a = 15.69

Substitute the given values into the formula and solve for A:

\implies \textsf{Area}=\sf \dfrac{8 \cdot 13 \cdot 15.69}{2}

\implies \textsf{Area}=\sf \dfrac{1631.76}{2}

\implies \textsf{Area}=\sf 815.88\:\:square \:units

<u>Area of the square</u>

\implies \textsf{Area}=\sf 13^2=169 \:\:square \:units

<u>Area of the shaded region</u>

= area of the octagon - area of the square

= 815.88 - 169

= 646.88

= 647 square units (nearest square unit)

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Thirty two packs can be made

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