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3 years ago

How to do the substitution in math?

Mathematics

1 answer:

seraphim [82]3 years ago

4 0

Substitution method can be applied in four steps

Step 1:

Solve one of the equations for either x = or y = .

Step 2:

Substitute the solution from step 1 into the other equation.

Step 3:

Solve this new equation.

Step 4:

Solve for the second variable.

hope this helps you!

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What is the GCF of two prime numbers?

ehidna [41]

Prime numbers: numbers which only have the factors of themselves and 1.

They include numbers such as 17, 19, 23, etc. These are numbers which only have two factors: such as 17, with the factors only being 17 and 1.

So therefore, this means, that for any two prime numbers, unless they are the same number, the only factor they will share is 1. So therefore the GCF of two prime numbers is 1.

Hope this helped!

4 0

3 years ago

Write a function rule for the relationship. the amount c of energy you burn and the time t you spend exercising, if you burn Cal

gregori [183]

Answer:

c= 12t

c is the amount of energy you burn, so it'll be the "solution" to the equation. Multiply 12 times the amount of time you spend exercising to find how much energy you'll burn.

6 0

3 years ago

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$