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2 years ago

На какой угол повернётся минутная стрелка за:30мин,15мин,10мин,1мин. Плиииииииииизззззззззззз и желательно быстрей

Mathematics

2 answers:

dexar [7]2 years ago

8 0

Just like she said
30 minutes= 180 degrees
15 minutes= 90 degrees
10 minutes= 60 degrees
1 minute = 6 degrees

jolli1 [7]2 years ago

5 0

30 mins = 180°
15 mins = 90°
10 mins = 60°
1 min = 6°

hope this helps you

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

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3 years ago

Which quadratic equation is equivalent to (x+2)^2+5(x+2)-6=0?

Mrrafil [7]

U=(x+2)
u^2+5u-6=0

C is answer

8 0

2 years ago

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What is the expression in radical form? (4x3y2)310

sertanlavr [38]

Given:

Consider the given expression is

$(4x^3y^2)^{\frac{3}{10}}$

To find:

The radical form of given expression.

Solution:

We have,

$(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

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3 years ago

How would you solve for the question below??? step by step please and thanks.

Airida [17]

Combine the two equations in the right amounts to eliminate y :

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x = 4

Solve for y :

2x + 3y = 17

8 + 3y = 17

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2 years ago

Find the following trigonometry values cos (150*) = ? sin (150*) = ?

melisa1 [442]

Answer:

See below in bold.

Step-by-step explanation:

cos 150 = - cos(180 - 150)

= - cos 30

= -√3/2.

sin 150

= sin (180 - 150

= sin 30

= 1/2.

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3 years ago

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