All categories

2 years ago

Need help with this please no spam

Mathematics

2 answers:

Nana76 [90]2 years ago

7 0

Answer:

9/2

Step-by-step explanation:

hope it's correct..,,,,,,,,,

Alona [7]2 years ago

6 0

Answer:

a. 9/2

Step-by-step explanation:

72:16

9:2

: ratio can be in a form of fraction as well

You might be interested in

What is the measure of angle A in the triangle below?

laiz [17]

The answer is 30.
You need to use sine here
So:

5 0

2 years ago

Find the area of the right triangle

xenn [34]

Answer:

Area: 24 sq units

Step-by-step explanation:

Area of a rectangle formula:

A = 0.5bh

Given:

b = 6

h = 8

Work:

A = 0.5bh

A = 0.5(6)(8)

A = 3(8)

A = 24

3 0

2 years ago

Read 2 more answers

What is the classification for this angle?

muminat

Answer:

The angle being shown is an acute angle

Step-by-step explanation:

7 0

1 year ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

2 years ago

Pls hurry it is 23 points i need help now Match the equation on the left with the correct estimate on the right. Each equation h

azamat

It looks floppy but you said you need this fast so yeah

6 0

2 years ago

Read 2 more answers