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3 years ago

Need help here plss

Mathematics

1 answer:

Ierofanga [76]3 years ago

6 0

Two linear equations can have no solutions, exactly one solution or infinitely many solutions. There will be no solution if the lines are parallel on a graph. There will be exactly one solution if the lines intersect each other on a single point. And finally, there will be infinite solutions if the lines overlap each other perfectly.

A single line however has infinite ordered pair solutions as the line travels infinitely in both directions on the coordinate plane. For example, using the equation y=3x, for any real value of x, we will get a real value for y.

Linear inequalities with two variables have infinitely many solutions. We can use the inequality y>3x as an example. For any real value of x, we will get a real value for y.

I hope this helps!

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The original purchase price of a car is 16,000, Each year, its value depreciates by 10%. Three years after its purchase, what is

Tju [1.3M]

The answer is 11,664

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3 years ago

Find Horizontal and Vertical asymptotes for y=(x^2 -9)/(4x^2 +1)

Gnom [1K]

Answer: Horizontal asymptote is $\dfrac{1}{4}$ and vertical asymptotes are $\pm \dfrac{1}{2i}$

Step-by-step explanation:

Since we have given that

$y=\dfrac{x^2-9}{4x^2+1}$

We need to find the horizontal and vertical asymptotes.

Since vertical asymptotes will occur where the denominator becomes zero.

So, here denominator is $4x^2+1$

Now,

$4x^2+1=0\\\\4x^2=-1\\\\x^2=\dfrac{-1}{4}\\\\x=\pm \dfrac{1}{2i}$

And the horizontal asympototes will occur when the coefficient of higher degree of numerator is divided by coefficient of higher degree of denominator.

$y=\dfrac{1}{4}$

Hence, horizontal asymptote is $\dfrac{1}{4}$ and vertical asymptotes are $\pm \dfrac{1}{2i}$

8 0

3 years ago

PLEASE HELP ME!!!

Harman [31]

Send a better photo or tell me what I’m suppose to be answering

8 0

3 years ago

the third term of an arithmetic sequence is 24 and the fifth is 32 if the first term is a1. which is an equation nth term of the

const2013 [10]

An = a1 + (n - 1)(d)
Where a1 is the first term and d is the common difference.
First find d, the common difference.
24, ____, 32
a3 a4 a5
Subtract 32-24 = 8
Subtract a5 - a3 = 2
Divide 8/2 = 4
d = 4
Use d and one of the values they give us to find a1.
a3 = 24
24 = a1 + (3 - 1)(4)
24 = a1 + 2(4)
24 = a1 + 8
Subtract 8 from both sides
16 = a1
an = 16 + (n - 1)(4)
Can also be written
an = 16 + 4n - 4
an = 4n + 12

8 0

3 years ago

Read 2 more answers

Help pls! find g(x)+ f(x)=-3x+4 and g(x)=5x+12

Igoryamba

(-3x+4) + (5x + 12) = 2x +16
The answer is D.
Hope it helps!

6 0

3 years ago

Read 2 more answers