Identify this conic section. 16y = x^2

What is the solution to the following system of equations?

kipiarov [429]

Answer:

(0, -6)

Step-by-step explanation:

Given the following systems of linear equations;

3x - 2y = 12 ...... equation 1

16x - 4y = 24 ........ equation 2

We would solve for the solution using the elimination method;

Multiplying eqn 1 by 2, we have;

2 * (3x - 2y = 12)

6x - 4y = 24

16x - 4y = 24

Subtracting the two equations, we have;

(6x - 16x) + (-4y -[-4y]) = (24 - 24)

-10x - 0 = 0

-10x = 0

x = -0/10 = 0

Next, we would find the value of y;

3x - 2y = 12

3(0) - 2y = 12

0 - 2y = 12

-2y = 12

y = -12/2

y = -6

Check:

3x - 2y = 12

3(0) - 2(-6) = 12

0 - (-12) = 12

12 = 12

Note: the options provided for this questions are incorrect or inappropriate.

5 0

3 years ago

What is 4/8 divided by 7/8

kifflom [539]

Answer: 4/7

Step-by-step explanation:

4/8 / 4/7

now you do multilpication so,

4x8

over

8x7

32/56 divided by 56/8=

4/7

hope i helped <33

8 0

3 years ago

Read 2 more answers

Find the product of 51 and -2.5.

Bad White [126]

Answer: C. -102 + (-25.5)

4 0

3 years ago

40% of what number is 28

Leto [7]

70
0.4x=28
x=28/0.4
x=70

7 0

3 years ago

Read 2 more answers

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

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

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

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago