What is the value of the expression (146.33-57)

I really need help with this question

Dominik [7]

Answer:

your screen is blue i can't see the problem dummy

5 0

3 years ago

Y=x +5 5x + 2y=1 Can you show the steps thanks

lubasha [3.4K]

Answer:

(-9/7,-26/7)

[more detailed at the bottom of the explanation]

Step-by-step explanation:

I am assuming that this is a system of equations...

So, knowing that y is equal to x + 5, you would plug that in the second equation and find x. Then you would plug x and find y.

—————

Step 1)

Equation 1:

y = (x + 5)

Equation 2:

5x + 2y = 1

Equation 2 can also be written as:

5x + 2(x+5) = 1

I wrote x+5 instead of y because the first equation tells us that y is equal to x + 5.

—————

Step 2)

Solve for x .

5x + 2(x+5) = 1

5x + 2x + 10 = 1

7x + 10 = 1

7x = 1 - 10

7x = -9

x = -9/7

—————

Step 3)

Plug the x value back in the first equation.

y = -9/7 + 5

y = -26/7

—————

Solution:

x = -9/7

y = -26/7

In ordered pair form:

(-9/7,-26/7)

————— ————— —————

Hope this helps !!!!!!!

5 0

3 years ago

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

Find the area of a circle who is diameter is 27 inches(use 3.1416)

Nimfa-mama [501]

Answer: The area is 572.5566

Step-by-step explanation:

27÷2=r=13.5

A=πr²

A=3.1416(π) · 13.5²(r)

A=572.5566

4 0