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2 years ago

5

3x = 2y. Is this equation a linear or non-linear relation? How do you know?

1 answer:

viktelen [127]2 years ago

7 0

Answer:

linear equation
all variables are to the first power

Step-by-step explanation:

"Linear equation" is another way to say "first-degree equation." That is, the equation involves constants and variables to the first degree only. No term contains any product or ratio of variables.

A term is first degree when the sum of all of the exponents of the variables is 1, and all exponents are positive integers. That is, the only allowed exponent of the single variable in a term is 1.

Here, the term on the left side of the equal sign has x to the first power. The term on the right side of the equal sign has y to the first power. Both are linear terms, so this is a linear equation.

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Simplify the polynomial. Then find its value for the given value of the variable.

vekshin1

Answer:

Step-by-step explanation:

Simplified expression:

$3x^3 -x^3 +5x^3-9x^3$

$2x^3 +5x^3-9x^3$

$7x^3 -9x^3$

$-2x^3$

If x = -1, the value of the expression is:

$-2x^3$

$-2(-1^3)$

$-2(-1)$

$2$

Hope this helps!

3 0

2 years ago

What angle corresponds to angle AFG?

viva [34]

Look at the image below :)

In case you didn't know, corresponding angles are equal to each other and are on the same side of the traversal

Therefore the answer is angle CGH

Hope this helped!

7 0

3 years ago

Read 2 more answers

Dahasolnce [82]

Answer:

just sub in the numbers into the equation to replace the X ..

example for

a) h(7) : 44

Step-by-step explanation:

h(7) : x^2 -5

sub in 7 into the equation to replace X

so you get

(7)^2 - 5

49-5

44

4 0

3 years ago

Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr

Bad White [126]

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:

$x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}$

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:

$x = \frac{-b +- R }{2*a}$

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

$x = \frac{-b +- C*i }{2*a}$

We have two complex solutions.

If D = 0

√0 = 0

then:

$x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}$

We have only one real solution (or two equal solutions, depending on how you see it)

3 0

3 years ago

I have no idea what this is since I was absent and my teacher is didn’t it explain it to me, someone?

MatroZZZ [7]

Answer:

Step-by-step explanation:

4 0

2 years ago