All categories

2 years ago

Ramon uses models to compare two decimial he says 0.09 greater than 0.1 because 9>1 which of these explain ramon's mistake

Mathematics

1 answer:

Shalnov [3]2 years ago

5 0

Answer:

Ramon's mistake was that he used whole numbers in comparison to decimals, which is not an accurate way to solve. Think about it this way, 0.1 is the tens place, 0.01 is the hundreds place. Which would be greater 100% of the time? The hundreds place, in decimal terms, the hundredths.

0.09 is greater than 0.1

You might be interested in

What is 3 divided by 0.05?

lilavasa [31]

Answer:

The answer is 60

6 0

3 years ago

Read 2 more answers

In QRS the measures of S = 90° and the measures of Q = 64° and RS = 16 feet find the length of QR to the nearest 10th of a foot

leva [86]

Answer:

vfd

Step-by-step explanation:

ddf

7 0

2 years ago

will giver brainliest to the person with the most detailed answer. That means tell me how you got ur answer. And the person who

Umnica [9.8K]

Answer:

She waited five minutes.

Step-by-step explanation:

You can tell that she waited five mins because between points D and E there is a straight line meaning that she was not moving. So you know that she was standing still waiting.

I hope this helps!!!

4 0

2 years ago

Read 2 more answers

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago

Choose a number between 67 and 113 that is a multiple of 2 6 and 10

Lilit [14]

Im going to go on and say that it is 90

8 0

3 years ago

Read 2 more answers