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3 years ago

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Mathematics

1 answer:

irina [24]3 years ago

5 0

Answer:

A cell wall is a structural layer surrounding some types of cells, just outside the cell membrane. It can be tough, flexible, and sometimes rigid. It provides the cell with both structural support and protection, and also acts as a filtering mechanism.

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The ratio of the number of Miki's stickers to the number of Ken's

Lera25 [3.4K]

Answer:

Miki had 288 stickers and Ken had 252 stickers.

Step-by-step explanation:

This question is solved using a system of equations.

I am going to say that:

Miki has x stickers.

Ken has y stickers.

The ratio of the number of Miki's stickers to the number of Ken's stickers was 8:7.

This means that $\frac{x}{y} = \frac{8}{7}$ , that is: $7x = 8y$ , or $x = \frac{8y}{7}$

After Miki gave Ken 18 of her stickers, they had the same number of stickers.

This means that:

$x - 18 = y + 18$

$x - y = 36$

Since $x = \frac{8y}{7}$

$\frac{8y}{7} - y = 36$

$\frac{8y}{7} - \frac{7y}{7} = 36$

$\frac{y}{7} = 36$

$y = 36*7 = 252$

And

$x - y = 36$

$x = 36 + y = 36 + 252 = 288$

Miki had 288 stickers and Ken had 252 stickers.

7 0

2 years ago

PLZ HELP URGENT!!!!!

Step2247 [10]

Answer:

A angle side angle

Step-by-step explanation:

<B = <D

Side BC = Side CD

<C = <C

We have angle side angle

4 0

3 years ago

Read 2 more answers

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

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

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

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

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$