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2 years ago

1. How many fifties are there in one hundred thousand?

Mathematics

1 answer:

VladimirAG [237]2 years ago

7 0

There are 2,000 fifties in one hundred thousand

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Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

3 0

3 years ago

Find what percent 14 is of 23 to the nearest whole percent?

Kazeer [188]

14 is .608, roundedto the nearest percent 14 is 61% of 23

4 0

3 years ago

3*(-4) really confuses need help

Arturiano [62]

Answer:

-12

Step-by-step explanation:

When you multiply a positive by a negative, the answer is always a negative. If you multiply two negatives, the answer is always positive. If you multiply 2 positives, the answer will always be positive.

7 0

3 years ago

What is 20:10 in its simplest<br><br> form

Hunter-Best [27]

Answer:

2/1

Step-by-step explanation:

20/10 devided by 10 to both numbers

3 0

3 years ago

To evaluate log2(3), Autumn reasoned that since log2(2) = 1 and log2(4) = 2, log2(3) must be the average of 1 and 2 and therefor

geniusboy [140]

Answer:

log₂(3) = 1.585 ≠ 1.5

Her thinking is not valid because the technique of average is valid only if the graph of the function is a straight line, but the graph of the log function is not a straight line.

Therefore the values cannot be taken by average

Step-by-step explanation:

Given:

log₂(2) = 1

log₂(4) = 2

To evaluate : log₂(3)

Now,

we know that

logₓ(y) = $\frac{\log(y)}{\log(x)}$ (Here the log has same base in the numerator and the denominator i.e 10)

therefore,

log₂(3) = $\frac{\log(3)}{\log(2)}$

also,

log(2) = 0.3010

log(3) = 0.4771

thus,

log₂(3) = $\frac{0.4771}{0.3010}$

log₂(3) = 1.585 ≠ 1.5

Her thinking is not valid because the technique of average is valid only if the graph of the function is a straight line, but the graph of the log function is not a straight line.

Therefore the values cannot be taken by average

7 0

3 years ago

1. How many fifties are there in one hundred thousand?​

1. How many fifties are there in one hundred thousand?