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3 years ago

What is the slope of the line y= 1

Mathematics

1 answer:

Harrizon [31]3 years ago

8 0

Answer:

The slope of y = 1 is 0

explanation:

Determining slope has been based on a simple formula, y2-y1 over x2-x1.

It is crucial to follow this formula in order to attain the proper slope.

When a simple expression is given, for instance; y = 1 then it could be divided by 0; thus, resulting in the answer simply being zero.

However, x=1 would contain the slope of undefined because any number divided by zero results in an undefined answer.

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Given y=-3/4x-2. What is the equation of the line that is perpendicular and has the point (-4,1)? Write answer in slope intercep

raketka [301]

Answer:

y=4/3+19/3

Step-by-step explanation:

When it asks for perpendicular it means your focusing on the slope.

Perpendicular means opposite of the current slope, so flip it and change the sign (-3/4 changes to 4/3). Next you will use the equation (y-y1)=m(x-x1)

y1=the y value given

x1=the x value given

m=your slope

(y-1)=4/3(x+4) Since the x point is minus a negative we get to change the sign to a positive. Next distribute the 4/3 to each term in the parenthesis.

y-1=4/3(x)+4/3(4)

y-1=4/3x+16/3 Now we add the one to both sides.

+1 +1 The one isn't in teh correct format to just add so first imagine it as 1/1. Now to get the bottom to have a 3 like in 16/3 we just multiply the top and bottom by 3.

$\frac{1}{1} x\frac{3}{3}=\frac{3}{3}$ Now we can add this to the 16/3

$\frac{16}{3} +\frac{3}{3}=\frac{19}{3}$ We can now put this back into our equation.

y=4/3+19/3

6 0

3 years ago

Need help answering this question please help

ohaa [14]

i think its C or D

yes

6 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$