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2 years ago

What point has the coordinates

Mathematics

2 answers:

Lana71 [14]2 years ago

7 0

The answer from me is M

Sloan [31]2 years ago

5 0

Answer:

Point M

Step-by-step explanation:

Since X is positive and Y being negative you would look in the bottom right quadrant

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What is the sum of the series?

Natalija [7]

Answer:

Step-by-step explanation:

k=1: 2*1-4 = -2

k=2: 2*2²-4 = 4

k=3: 2*3²-4 = 14

k=4: 2*4²-4 = 28

-2 + 4 + 14 + 28 = 44

4 0

3 years ago

Read 2 more answers

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

6 0

2 years ago

Read 2 more answers

Suppose three riders rode a total of 182 miles. If they used a total of 13 horses, and rode each horse the same number of miles,

umka21 [38]

Answer:

They rode 14 miles before replacing each horse

Step-by-step explanation:

We will be working under the assumption that all three riders sat on one horse at a time and rode it while the other horses rested.

From the problem, we can understand that the horses were each ridden for the same distance. This means that to get the total distance a horse rode before it was changed, we can divide the total distance by the number of horses that were used for the journey.

Distance each horse rode = 182/ 13 = 14 miles.

Therefore, each horse was ridden for 14 miles before it was changed.

5 0

2 years ago

Please help me find the area and perimeter of this composite figure.

pentagon [3]

Answer:

the area is 26 the perimetier is 52

Step-by-step explanation:

4 0

2 years ago

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Last math question for today!!!!

Alex777 [14]

Answer: Maybe search it up?...I cant rlly see what the question says

Step-by-step explanation:

3 0

2 years ago