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3 years ago

Will mark BRAINLIEST!!!

Mathematics

2 answers:

sergiy2304 [10]3 years ago

8 0

Your answer should be c ....................

guapka [62]3 years ago

8 0

Answer:

$\frac{\sqrt[4]{3} }{4xy^{2} }$ Your answer should be C.

Step-by-step explanation:

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A commercial building contractor is trying to decide which of two projects to commit her company to. Project A will yield a prof

Alexxandr [17]

Answer:

Project A: $55,000 Project B: $50,800 Contractor should choose project A

Step-by-step explanation:

The expected value of the project is the sum of products of profit and its probability:

In thousands, ...

A: (50×0.6) +(80×0.3) +(10×0.1) = 30 +24 +1 = 55 . . . thousand

B: (100×0.1) +(64×0.7) +(-20×0.2) = 10 +44.8 -4 = 50.8 . . . thousand

The contractor should choose Project A based on the expected value of profit.

3 0

3 years ago

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

Correct the error. What should the line actually be?

Natali5045456 [20]

Answer:

A line....

Step-by-step explanation:

Plz be more clear with your question!

Thank You

3 0

3 years ago

Write an expression, using an exponent, that is equivalent to 9\times 9\times 9\times 9\times 9\times 9\times 99×9×9×9×9×9×9

jeka57 [31]

Nine to the eighth power

7 0

3 years ago

Read 2 more answers

Consider M, N, and P. collinear points on MP.

Kobotan [32]

Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}$

2. Point N lies to the right from point P, then MP + PN = MN. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}$

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}$

3 0

3 years ago