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3 years ago

Solve for Xin the equation x^2-12+36=90

Mathematics

2 answers:

ivann1987 [24]3 years ago

8 0

Answer:

12+36=90

Step-by-step explanation:

(3/5+7i)+(0.3-7.002i)

(5+77i)-(19/4-8i)

(240-22i)*(247/7+2.222i)

(33/5+99i)/(33/5-88i)

Glad i could help!

Hunter-Best [27]3 years ago

7 0

X= + or- square root 66 i solved using the pq formula but u can also use the quadratic formula as well and the square root formula but I'ma show u how to solve it with pq it's on this image hope it helps and mark me brainliest

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Solve the following System of Three Equations: x−3y+z=−15<br> 2x+y−z=−2<br> x+y+2z=1

jeyben [28]

Answer:

x = -3 , y = 4 , z = 0

Step-by-step explanation:

Solve the following system:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for z:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Solve for z.

Subtract x - 3 y from both sides:

{z = 3 y + (-x - 15)

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Perform a substitution.

Substitute z = -15 - x + 3 y into the second and third equations:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

x + y + 2 (-15 - x + 3 y) = 1

Hint: | Expand the left hand side of the equation x + y + 2 (-15 - x + 3 y) = 1.

x + y + 2 (-15 - x + 3 y) = x + y + (-30 - 2 x + 6 y) = -30 - x + 7 y:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Isolate terms with x to the left hand side.

Subtract 15 - 2 y from both sides:

{z = -15 - x + 3 y

3 x = 2 y - 17

-30 - x + 7 y = 1

Hint: | Solve for x.

Divide both sides by 3:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

-30 - x + 7 y = 1

Hint: | Perform a substitution.

Substitute x = (2 y)/3 - 17/3 into the third equation:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Choose an equation and a variable to solve for.

In the third equation, look to solve for y:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Isolate terms with y to the left hand side.

Add 73/3 to both sides:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 = 76/3

Hint: | Solve for y.

Multiply both sides by 3/19:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

y = 4

Hint: | Perform a back substitution.

Substitute y = 4 into the first and second equations:

{z = -x - 3

x = -3

y = 4

Hint: | Perform a back substitution.

Substitute x = -3 into the first equation:

{z = 0

x = -3

y = 4

Hint: | Sort results.

Collect results in alphabetical order:

Answer: {x = -3 , y = 4 , z = 0

6 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

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3 years ago

Solve the following system of equations by elimination. What is the value of y?

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