All categories

3 years ago

Solve and Graph -x+4y=-16

Mathematics

1 answer:

nikdorinn [45]3 years ago

3 0

Hoped that this helpedd!!

You might be interested in

575 394 9550 <br> bFzN0S <br> come study with me!

Molodets [167]

Answer:

need points

Step-by-step explanation:

gchfyfyjfuf

5 0

2 years ago

What is the change from -9 to -1?.

jarptica [38.1K]

Answer:

Positive 8

Step-by-step explanation:

Since they're both negative numbers and it went from -9 to -1, you had to have added a positive number to get a lesser negative number. To put it simply, if you were to subtract 8 from 9, you'd get 1.

5 0

3 years ago

Solve for x 21 27 x-3 x-1

Nataly_w [17]

Answer

27x^22-3x-1

hope this is right!!

Step-by-step explanation:

7 0

2 years ago

Using the binomial theorem, expand the expression (4 + y) to the power of 4

Flauer [41]

Answer:

When raised to the power of 4, the binomial (4 + y) expands to:

$y^4 + 16y^3 + 96y^2 + 256y + 256$

Step-by-step explanation:

$(4 + y)^4 \\= (y^2 + 8y + 16)^2\\= y^4 + 8y^3 + 16y^2 + 8y^3 + 64y^2 + 128y + 16y^2 + 128y + 256\\= y^4 + 16y^3 + 96y^2 + 256y + 256$

8 0

2 years ago

Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i

lana66690 [7]

In the $x,y$ plane, we have $z=0$ everywhere. So in the equation of the sphere, we have

$25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2$

which is a circle centered at (2, -10, 0) of radius 4.

In the $x,z$ plane, we have $y=0$ , which gives

$25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75$

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the $y,z$ plane, $x=0$ , so

$25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2$

which is a circle centered at (0, -10, 3) of radius $\sqrt{21}$ .

3 0

3 years ago