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3 years ago

What does it mean when ph is slightly acidic?

Mathematics

1 answer:

Nutka1998 [239]3 years ago

8 0

D. Slightly acidic is 5.0 to 5.5

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Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = cos(θ) + sin(10

vivado [14]

Answer:

Step-by-step explanation:

x = cos θ + sin(10θ)

y = sin θ + cos(10θ)

Take derivative with respect to θ:

dx/dθ = -sin θ + 10 cos(10θ)

dy/dθ = cos θ - 10 sin(10θ)

Divide:

dy/dx = (dy/dθ) / (dx/dθ)

dy/dx = (cos θ - 10 sin(10θ)) / (-sin θ + 10 cos(10θ))

Evaluate the derivative at θ=0:

dy/dx = (cos 0 - 10 sin 0) / (-sin 0 + 10 cos 0)

dy/dx = 1/10

Evaluate the parametric functions at θ=0:

x = cos 0 + sin 0 = 1

y = sin 0 + cos 0 = 1

Writing the equation of the tangent line in point-slope form:

y - 1 = 1/10 (x - 1)

6 0

3 years ago

What is the inverse of 4 square

Scrat [10]

$a^{-1}=\dfrac{1}{a}-\text{the inverse of a}\\\\\text{the inverse of}\ 4^2:\\\\\left(4^2\right)^{-1}=4^{-2}=\dfrac{1}{4^2}=\dfrac{1}{16}$

5 0

3 years ago

Can you make 24 from the numbers 3, 6, 8 and 2, using all the numbers? You can only use the operations: + - x / ( )

Andre45 [30]

Answer:

you can do 3 x 6 + 8 - 2

Step-by-step explanation:

3 × 6= 18

+8= 26

-2= 24

hope this Helps :)

4 0

2 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

Another name for segments qr

Helga [31]

Answer: segment rq

<u>Explanation:</u>

Segments are named by their endpoints. You can start/end with either endpoint. So, qr can also be called rq.

7 0

2 years ago