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3 years ago

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Mathematics

2 answers:

vampirchik [111]3 years ago

8 0

X=7 this is what i got trying to solve this

melisa1 [442]3 years ago

4 0

Answer:

Maple calculator used_____

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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

Which expression is equivalent to j (j) (j) (j) (j) (j) (j) (j) (j) (j) (j) (j) (j)?

Ahat [919]

Answer:

j^13

Step-by-step explanation:

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3 years ago

2(5m^4n^-1)^3 simplify show work please

dangina [55]

Answer:

$\large\boxed{2(5m^4n^{-1})^3=250m^{12}n^{-3}=\dfrac{250m^{12}}{n^3}}$

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\-------------------\\\\2(5m^4n^{-1})^3=2(5)^3(m^4)^3(n^{-1})^3=2(125)m^{3\cdot4}n^{-1\cdot3}=250m^{12}n^{-3}=\dfrac{250m^{12}}{n^3}$

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3 years ago

What is the derivative of the function below?

11111nata11111 [884]

$g(x)=4x-\dfrac{1}{4x}=4x-(4x)^{-1}\\\\\\\\g'(x)=\left[4x-(4x)^{-1}\right]'=(4x)'-\left[(4x)^{-1}\right]'=4-[-1(4x)^{-2}]\cdot4\\\\=4+\dfrac{1}{(4x)^2}\cdot4=4+\dfrac{4}{4^2x^2}=\boxed{4+\dfrac{1}{4x^2}}\\\\Used:\\\\(f(x)-g(x))'=f'(x)-g'(x)\\\\.[f(g(x))]'=f'(g(x))\cdot g'(x)\\\\(x^n)'=nx^{n-1}$

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3 years ago

After a storm damages the community center, Shanika and her friends hold fundraising events to help pay for repairs. After the f

pantera1 [17]

Answer:

Step-by-step explanation:

The total they want to raise is $2,400 if that is what you're asking?

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2 years ago