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nikdorinn [45]
3 years ago
15

The path of a race will be drawn on a coordinate grid like the one shown below. The starting point of the race will be at (−5.3,

1). The finishing point will be at (1, −5.3).
A four-quadrant coordinate grid is drawn from negative 4 to positive 4. Quadrant I is labeled Quadrant P, Quadrant II is labeled Quadrant Q, Quadrant III is labeled Quadrant R, and Quadrant IV is labeled Quadrant S.

Part A: Use the grid to determine in which quadrants the starting point and the finishing point are located. Explain how you determined the locations. (6 points)

Part B: A checkpoint will be at (5.3, 1). In at least two sentences, describe the difference between the coordinates of the starting point and the checkpoint, and explain how the points are related. (4 points)

Mathematics
2 answers:
DochEvi [55]3 years ago
8 0

Answer:

a

Step-by-step explanation:

kompoz [17]3 years ago
8 0

Answer:

Part A:  For the -5.3, 1 it would be located in Quadrant Q because even though it just goes to -4 and 4 I can extend the grid in my head and that's how I was able to figure out.

Part B:   From the origin (0,0) you will go 5 spaces to the left and then a little more to the left like in the middle of 6 and 5 but more towards the 5.

Step-by-step explanation:

I'm BIG BRAIN yessir SHEEEEEEESH

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Https://www.desmos.com/calculator use this to graph your equations like this, it is very helpful to solve difficult math problems like these. :) The vertex is (-4,1) and even shows it on the graph
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What's (6x+8)(5x-1)<br> Can you help me solve it please
11111nata11111 [884]
(6x+8)(5x-1)=30x^2-6x+40x-8=30x^2+34x-8
3 0
3 years ago
Help please.... 14 points..?
-Dominant- [34]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
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Hi can anyone help me with this please ​
kupik [55]

Answer:

sure

Step-by-step explanation:

she multiplied it 3 times, so, i dont think 8 should be negative.

6 0
3 years ago
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Which would be the best method to use to solve the following equations? Explain your reasoning.
o-na [289]

A. square root property

3x^2-192=0\\x^2-66=0\\x^2=66\\x= -\sqrt{66} or \sqrt{66}

it has one value with x which is x^2 and it cn be easily solved without having to factorise, quadratic formula cant be used as it need ax^2+bx+c=0 format

B. factorising

x^2-x-6=0\\x^2+2x-3x-6=0\\(x+2)(x-3)=0\\x+2=0  and  x-3=0\\x= -2 \\x= 3

i just felt like this was easier to factorise than the other 2 options left

C. Completing the square

x^2-6x-7=0\\(x-3)^2=0\\x= -1\\x= 7

same reason personal preference

D- Quadratic

x^2-17x-7=0\\x=\frac{-(-17)+\sqrt{(-17)^2-4(1)(-7)} }{2(1)} \\x=\frac{-(-17)-\sqrt{(-17)^2-4(1)(-7)} }{2(1)}\\x=17.4\\x=-0.402

the 17 kinda threw me off and i didnt wanna get on factorising or doing completing the square so quadratic formal

7 0
3 years ago
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