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2 years ago

13. What point is in the third quadrant? a. (1,-2) b. (-2,-1) C (-2.1) d (1.2)

Mathematics

2 answers:

Murrr4er [49]2 years ago

8 0

The third quadrant are both negative numbers so the answer is B.

WARRIOR [948]2 years ago

7 0

I believe the answer will be b. (-2,-1). good luck.

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Factor the binomial using its GCF: 10x-15x^4

Igoryamba

Answer:

5x(2 - 3x³)

Step-by-step explanation:

10x-15x^4

The GCF( Greatest Common Factor) of 10x-15x^4 is 5x

Solve:

10x/5x = 2

-15^4/5x = -3x³

Final answer is

5x(2 - 3x³)

<u>Check</u>

5x(2 - 3x³)

distribute

5x(2) + 5x(-3x³)

10x - 15x^4

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3 years ago

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What is the ratio of the area of the inner square to the area of the outer square?

Anastasy [175]

Answer:

$\frac{(a-b)^2+b^2}{a^2}$

Step-by-step explanation:

Since, By the given diagram,

The side of the inner square = Distance between the points (0,b) and (a-b,0)

$=\sqrt{(a-b-0)^2+(0-b)^2}$

$=\sqrt{(a-b)^2+b^2}$

Thus the area of the inner square = (side)²

$=(\sqrt{(a-b)^2+b^2})^2$

$=(a-b)^2+b^2\text{ square cm}$

Now, the side of the outer square = Distance between the points (0,0) and (a,0),

$=\sqrt{(a-0)^2+0^2}$

$=\sqrt{a^2}=a$

Thus, the area of the outer square = (side)²

$=a^2\text{ square cm}$

Hence, the ratio of the area of the inner square to the area of the outer square

$=\frac{(a-b)^2+b^2}{a^2}$

4 0

3 years ago

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A passenger rode the subway 2 blocks west and then 10 south. If all the blocks are the same length what is the value of the slop

Nimfa-mama [501]

Answer:

The slope of the line is 5

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

I have created an illustration which is attached below in order to help you understand the situation. As we are told the passenger rode the subway from point (0 , 0) to point (-2, 0). Then he rode the subway from point (-2, 0) to the final destination of point (-2 , -10). In order to find the value of the slope we need to use the slope formula which is

$Slope = \frac{y_{2}-y_{1} }{x_{2}- x_{1} }$