All categories

3 years ago

The size of a cell is typically found by capturing an image under a microscope then using software to measure its diameter.

Mathematics

2 answers:

trasher [3.6K]3 years ago

8 0

Answer:

8.95 x 10^-3

kakanansjammz

Margarita [4]3 years ago

3 0

Answer:

Step-by-step explanation:

The answer is 6.868 10

You might be interested in

What is (5 radical 2) squared?

Roman55 [17]

(5 radical 2)^2 =

5^2 × (radical 2)^2 =

25 × 2 = 50

3 0

3 years ago

Bertha and Vernon are competing in a diving competition. Bertha's dive ended -54 m from the starting platform. Vernon's dive end

irinina [24]

.bertha dived 8 times more then vernon 54/8 = 6

8 0

3 years ago

Help math homework both questions

-Dominant- [34]

Answer:

Q-12 The vertical length of the top part of elevator to ground is 0.8484 meters .

Q -14 The glide runway path is 24,752.47 feet .

Step-by-step explanation:

Q - 12

Given as :

The diagonal length of the elevator = e = 1.2 meters

The vertical length of the top part of elevator to ground = x meters

The angle made by elevator with ground = Ф = 45°

Now, According to figure

Sin angle = $\dfrac{\textrm perpendicular}{\textrm hypotenuse}$

Or, Sin Ф = $\dfrac{\textrm x}{\textrm e}$

Or, Sin 45° = $\dfrac{\textrm x meters}{\textrm 1.2 meters}$

Or, 0.707 × 1.2 meters = x

∴ x = 0.8484 meters

So, The vertical length of the top part of elevator to ground = x = 0.8484 meters

Hence,The vertical length of the top part of elevator to ground is 0.8484 meters . Answer

Q-14

Given as:

The altitude of decent of plane = H = 10,000 feet

The angle of descend = Ф = 22°

Let the glide runway path = y feet

Now, According to question

Tan angle = $\dfrac{\textrm perpendicular}{\textrm base}$

Or, Tan Ф = $\dfrac{\textrm H}{\textrm y}$

Or, Tan 22° = $\dfrac{\textrm 10,000 feet}{\textrm y feet}$

Or, 0.4040 = $\dfrac{\textrm 10,000 feet}{\textrm y feet}$

Or, y = $\dfrac{\textrm 10,000 feet}{\textrm 0.4040}$

∴ y = 24,752.47

So,The glide runway path = y = 24,752.47 feet

Hence, The glide runway path is 24,752.47 feet . Answer

6 0

3 years ago

Do u need a common denominator if you have adding and dividing in fractions?

Rama09 [41]

You only need a common denominator for adding and subtracting!

4 0

3 years ago

Read 2 more answers

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago