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2 years ago

A full bag of dog food weighs 512 pounds.

Mathematics

2 answers:

uysha [10]2 years ago

4 0

None of them is the correct answer

Iteru [2.4K]2 years ago

3 0

Answer:

umm

Step-by-step explanation:

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What is the value of 4/15 divided by 2/3

ruslelena [56]

Answer:

0.5

Step-by-step explanation:

I'm guessing os yeahhh

6 0

3 years ago

What is the length of AD if A is located at (-7,0) and D is located at (8,0)?

Kryger [21]

Answer:

Step-by-step explanation:

(-7,0) and (8,0) are on the opposite side of origin of x axis

d = 8 - -7 = 15

6 0

3 years ago

Choose the best score on a quiz 15 points out of 25

Nezavi [6.7K]

15 out of 25 is a 60%

6 0

2 years ago

Pls say hard very hard

Simora [160]

(a) From the histogram, you can see that there are 2 students with scores between 50 and 60; 3 between 60 and 70; 7 between 70 and 80; 9 between 80 and 90; and 1 between 90 and 100. So there are a total of 2 + 3 + 7 + 9 + 1 = 22 students.

(b) This is entirely up to whoever constructed the histogram to begin with... It's ambiguous as to which of the groups contains students with a score of exactly 60 - are they placed in the 50-60 group, or in the 60-70 group?

On the other hand, if a student gets a score of 100, then they would certainly be put in the 90-100 group. So for the sake of consistency, you should probably assume that the groups are assigned as follows:

50 ≤ score ≤ 60 ==> 50-60

60 < score ≤ 70 ==> 60-70

70 < score ≤ 80 ==> 70-80

80 < score ≤ 90 ==> 80-90

90 < score ≤ 100 ==> 90-100

Then a student who scored a 60 should be added to the 50-60 category.

8 0

2 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