All categories

2 years ago

GIVING BRAINLIEST PLZZ HELP ASAP

Mathematics

2 answers:

stealth61 [152]2 years ago

7 0

Answer:

Step-by-step explanation:

x and y go up at a constant rate (or you can say it has a CROC or constant rate of chacge) y going up 5 for every one x

with our equation being y = 5x + 5

hope this helps <3

muminat2 years ago

5 0

Answer:

Step-by-step explanation:

It would be D because to know if its linear, x has to increase by 1 and y by a constant rate.

X increases by 1 and y increases by 5

I've attached a graph for you to see

If you have any further questions, don't be afraid to reach out! I hope this helps!

You might be interested in

3.75x-5-8.75X+4.5+0.5

exis [7]

Step-by-step explanation:

3.75x-5-8.75x+4.5+0.5

=3.75x-8.75x+4.5+0.5-5

=-5x+5-5

==-5x

4 0

2 years ago

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$

Substitute:

$\displaystyle y\left(\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\right) + \left(\frac{3e^{2x}-2}{y}\right)^2 =6e^{2x}$

Simplify:

$\displaystyle \frac{6y^2e^{2x}- \left(3e^{2x} -2\right)^2}{y^2} + \frac{\left(3e^{2x}-2\right)^2}{y^2}= 6e^{2x}$

Combine fractions:

$\displaystyle \frac{\left(6y^2e^{2x}-\left(3e^{2x} - 2\right)^2\right) +\left(\left(3e^{2x}-2\right)^2\right)}{y^2} = 6e^{2x}$

Simplify:

$\displaystyle \frac{6y^2e^{2x}}{y^2} = 6e^{2x}$

Simplify:

$6e^{2x} \stackrel{\checkmark}{=} 6e^{2x}$

Q.E.D.

6 0

2 years ago

4(x+3)=2(2x+9)

kramer

Answer:

m a t h w a y

Step-by-step explanation:

;)

7 0

2 years ago

Which ordered pair identifies a point in Quadrant IV? A) (2, 5) B) (0, 1) C) (-1, -1) D) (12, -5)

MariettaO [177]

Correct Answer:
Option D (12, -5)

In Quadrant IV, the x-component of the point is positive and the y-component is negative. From the given points, option D contains the point with a positive x-component i.e.12 and negative y-component i.e -5. Therefore, this point lies in Quadrant IV.

4 0

3 years ago

Which statement is true?

Mumz [18]

The answer is D.

|-4| < 4
4 < 4

|-1| < 0
1 < 0

|-6| = -6
6 = -6

|-14| > 10
14 > 10

6 0

2 years ago