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3 years ago

Who wanna be friends?

Mathematics

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:

yessssssss

Step-by-step explanation:

zhannawk [14.2K]3 years ago

5 0

Answer:

okkkkkkkkkkkkkkkkkkkkkk

Step-by-step explanation:

You might be interested in

Convert 65 millimeters to centimeters Pls use step by step pls

Blababa [14]

Answer:

65 mili= 6.5cm

Step-by-step explanation:

divide the length value by 10

6 0

2 years ago

Read 2 more answers

By selling a watch in Rs 150 and Rs 200 loss and profit are happened

Travka [436]

Answer:

Rs 175

Step-by-step explanation:

Suppose the cost is x and at Rs150 the loss is 150-x (this should be a negative number).

At Rs200, the profit is 200-x.

So we have an equation: minus 150 minus x is equal to 200 minus x.

To solve the equation, the cost price X is Rs175.

3 0

3 years ago

Kay is a student in Mrs. Hudson’s class. Assuming you know nothing else about Kay, what is the probability that Kay’s birthday w

Drupady [299]

Answer:

262/365

Step-by-step explanation:

So as you can see there is no more information aout Kay on her birthday, so the chances of her birthday being on a week day is given by the total number of the weekdays of the year between the total number of days in a year, so in 2019 there are 262 weekdays, divided by 365 you get the probability that Kay´s birthday falls on a weekday.

262/365=,7178=71,78%

So the probability of Kay´s brithday falling on a week day will be 71,72%

5 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$