Using simultaneous equation,
a=-1, b=1, x=1, y=0
OR
a=1, b=-1, x=1, y=0.
Answer:
A quadratic equation can be written as:
a*x^2 + b*x + c = 0.
where a, b and c are real numbers.
The solutions of this equation can be found by the equation:

Where the determinant is D = b^2 - 4*a*c.
Now, if D>0
we have the square root of a positive number, which will be equal to a real number.
√D = R
then the solutions are:

Where each sign of R is a different solution for the equation.
If D< 0, we have the square root of a negative number, then we have a complex component:
√D = i*R

We have two complex solutions.
If D = 0
√0 = 0
then:

We have only one real solution (or two equal solutions, depending on how you see it)
Your number is 0, 4x6=24 and six less than six is 0
Answer:

Step-by-step explanation:
1. Swap sides

Swap sides:

2. Isolate the y

Multiply to both sides by 18:

Group like terms:

Simplify the fraction:

Multiply the fractions:

Simplify the arithmetic:

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Why learn this:
- Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends?
- Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives.
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Terms and topics
- Linear equations with one unknown
The main application of linear equations is solving problems in which an unknown variable, usually (but not always) x, is dependent on a known constant.
We solve linear equations by isolating the unknown variable on one side of the equation and simplifying the rest of the equation. When simplifying, anything that is done to one side of the equation must also be done to the other.
An equation of:

in which
and
are the constants and
is the unknown variable, is a typical linear equation with one unknown. To solve for
in this example, we would first isolate it by subtracting
from both sides of the equation. We would then divide both sides of the equation by
resulting in an answer of:

Answer: t = 6 s
Step-by-step explanation:
When Rose hit the water, the height of the jump is 0, so: h(t) = 0.
Resolving it, we have:
-16t² + 16t + 480 = 0
To simplify the second degree equation, divided it by 16:
-t² + t + 30 = 0
Now, solve the equation using Bhaskara:
Δ = 1² - 4. (-1) . 30
Δ = 1 + 120
Δ = 121
t = 
t₁ =
= - 5
t₂ =
= 6
Since time can be negative, it took t = 6s for Rose to hit the water.