All categories

3 years ago

What is the equation of a line that is perpendicular to the line y=3x-6 and passes through the point (15,5)?

Mathematics

1 answer:

Goshia [24]3 years ago

3 0

the fouth option is correct ans....

You might be interested in

What is the difference between Delta x and dx? ...?

MA_775_DIABLO [31]

Δx (delta x) is about a secant line, a line between two points representing the rate of change between those two points. That's a "differential" (between the two points).

dx is about a tangent line to one point, representing an instantaneous rate of change. That makes it a "derivative."

6 0

3 years ago

A sunflower is 15 inches tall and grows 2 inches a week. How many weeks will it take for the

Zarrin [17]

Answer:

4 weeks

Step-by-step explanation:

15 + 2x = 23

2x = 8

x = 4

8 0

2 years ago

The nature park has a pride of 7 adult lions and 4 cubs. The adults eat 6 pounds of meat each day and the cubs eat 3 pounds. Sim

Mademuasel [1]

Answer:

7 x 6 = 42, 42+ 4 x 3, 4 x 3 is 12, 42 + 12 = 54

Step-by-step explanation:

Hope I helped pls give brainliest

8 0

2 years ago

How do I uses these

andrezito [222]

Answer:

um i dont know

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

4 years ago