All categories

3 years ago

Need help on this asap

Mathematics

1 answer:

Ganezh [65]3 years ago

5 0

(-5)(10)= 50 degrees

Brainliest?

You might be interested in

Find the compound interest on 800 at 5% per annum for 2 years.

Ludmilka [50]

Answer:

$C = 82$

Step-by-step explanation:

Given

$R = 5\%$ -- Rate

$P = 800$ -- Principal

$N = 2$ -- Time

Required

Determine the compound interest

First, we calculate the Amount (A)

$A = P(1 + R)^N$

$A = 800*(1 + 5\%)^2$

Express % as decimal

$A = 800*(1 + 0.05)^2$

$A = 800*(1.05)^2$

$A = 882$

The compound interest (C) is then calculated as:

$C = A -P$ i.e. Amount - Principal

$C = 882 - 800$

$C = 82$

7 0

2 years ago

It takes 2 minutes to fill a rectangular aquarium 8 inches long, 9 inches wide, 13 inches tall. How long will it take the same h

spin [16.1K]

Answer:

46.74 minutes

4 0

2 years ago

What is the factored form of 64g^3 +8?<br>

Gelneren [198K]

Answer:

8(2g + 1) (4g^2 - 2g + 1)

Step-by-step explanation:

8 0

3 years ago

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

3 years ago

A fair coin is flipped 5 times

wolverine [178]

Answer:

2/5

Step-by-step explanation:

The answer is common sense.

4 0

3 years ago