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IrinaK [193]
3 years ago
15

PLZ PLZ PLZ HELPPPP

Mathematics
1 answer:
Andrews [41]3 years ago
5 0

Answer:

LCD is 6

first choice

Step-by-step explanation:

least common denominator of 2 and 3 is 6

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These operation are called reverse operation......... I hope this helps 
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The equation of state of a certain gas is given by p = rt/vm + (a + bt)/vm2, where a and b are constants. find (∂v/∂t)p.
kotegsom [21]
Given:
p= \frac{RT}{vm} + \frac{a+bT}{vm^{2}} \\or\\v= \frac{1}{p}[ \frac{R}{m}  + \frac{a+bT}{m^{2}}]

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When p is held constant,
\frac{\partial v}{\partial T} |_{p}= \frac{1}{p}[ \frac{R}{m}+ \frac{b}{m^{2}}]
5 0
3 years ago
Ratio of 20 to24 is ​
Sedaia [141]

Answer:

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20:24

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6 0
3 years ago
Someone help pretty please!
Nonamiya [84]
The answer is choice B
y > (2/3)x + 1

The boundary line is the equation y = (2/3)x + 1 which can be found through the slope formula to get m = 2/3. Then you use one of the two points on the line to find b = 1. 

The equal sign in y = (2/3)x + 1 changes to a "greater than" sign to indicate two things
A) The shaded region is above the boundary line
B) The boundary line itself is a dashed line to indicate "no solution points on this line"
4 0
3 years ago
You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o
anyanavicka [17]

Answer:

x =\dfrac{45 \sqrt{6}}{ 2}

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}

In order to estimate the point along the shore, x meters from B, the runner should  stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

=\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0

-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0

\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}

squaring both sides; we get

\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}

\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}

By cross multiplying; we get

49x^2 = 25(54^2+x^2)

49x^2 = 25 \times 54^2+ 25x^2

49x^2-25x^2 = 25 \times 54^2

24x^2 = 25 \times 54^2

x^2 = \dfrac{25 \times 54^2}{24}

x =\sqrt{ \dfrac{25 \times 54^2}{24}}

x =\dfrac{5 \times 54}{\sqrt{24}}

x =\dfrac{270}{\sqrt{4 \times 6}}

x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}

x =\dfrac{45 \sqrt{6}}{ 2}

8 0
3 years ago
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