All categories

2 years ago

I really need help me find A×B when

Mathematics

1 answer:

lilavasa [31]2 years ago

5 0

Answer:

A=1,2,3

B=x^2 is the answer

You might be interested in

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

3 years ago

Read 2 more answers

What is the surface area of the regular pyramid given below?

yuradex [85]

For this case we have by definition, that the total surface area of a regular pyramid with a square base is given by:

$SA = \frac {1} {2} p * S + A_ {b}$

Where:

p: It is the perimeter of the base

S: It's the inclination

$A_ {b}:$ It is the area of the base

Substituting:

$A_ {b} = 7 ^ 2 = 49units ^ 2\\p = 7 + 7 + 7 + 7 = 28units\\S = 10units\\SA = \frac {1} {2} 28 * 10 + 49\\SA = 140 + 49\\SA = 189 \ units ^ 2$

ANswer:

Option B

3 0

3 years ago

Find the area of the trapezoid.

vlabodo [156]

Answer: A.90 cm^2

Step-by-step explanation: A= a+b/2 h

A= 17.5+12.5/2 *6

A = 30/2*6

A= 15*6

A=90

5 0

2 years ago

PLEASE HELP DUE SOON

Mazyrski [523]

Answer:

Step-by-step explanation:

The lines cross or touch the X-Axis 4 times which means that there are 4 solutions.

7 0

2 years ago

Can someone help me ASAP!!

Lana71 [14]

The answer is 5.

Step-by-step explanation:

To find the interquartile range you must look at the median and the upper and lower halves of the data. The median (12) to the upper half of the data (15) is 3. Next, look at the median (12) to the lower half of the data (10) it is 2. We then add 2 and 3 to get the interquartile range of 5.

6 0

2 years ago

I really need help me find A×B when​

I really need help me find A×B when