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3 years ago

the sale price of a computer is $700, which is 20% less than last month's price. what was laat month's price?

Mathematics

2 answers:

erica [24]3 years ago

8 0

Last months price was 3500

kogti [31]3 years ago

7 0

Answer

Step-by-step explanation:700 x .20 = 140. Then 700 - 140 =

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What is the greatest common factor of 27,45,63 and then 24 40 64

Aleks [24]

Answer:

Step-by-step explanation:

5 0

3 years ago

Ticket for football final selling well, on Thursday 4/7 of the tickets are sold, on Friday 1/4 of the tickets are sold, What fra

sergij07 [2.7K]

Answer:

5/28

Step-by-step explanation:

1/1 represents the whole amount of tickets.

4/7 are sold on Thursday.

an additional 1/4 are did on Friday.

fraction of available tickets on Saturday are therefore

1/1 - 4/7 - 1/4 = 4/1 - 16/7 - 1/1 = 28/1 - 16/1 - 7/1 = 5/1

that means out of 28 equal parts of the total number of tickets 16+7 = 23 parts have been sold.

so, there are only 5 out of the total 28 parts available on Saturday, and the fraction of available tickets is therefore 5/28.

we could have also tried to bring both fractions to the same denominator, and found that this is 28.

and then

4/7 + 1/4 = 4/4 × 4/7 + 7/7 × 1/4 =

= 16/28 + 7/28 = 23/28

and the whole is 28/28.

so we have 28/27 - 23/28 = 5/28 left.

surprise, surprise, it is the same result ...

4 0

2 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,

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

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

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

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

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

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

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

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

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago

Some help me please :( I really need help with this

katrin2010 [14]

First part, in quadrilateral WXYZ, WX is parallel to ZY because both sides are equal while the other parallel segments WZ and XY are equal but different side lengths.

The measure of angle Z is 78 because both of the pictures are the same, only rotated. and angle S and Z are acute.

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3 years ago

DELETED DELETED DELETED

Viefleur [7K]

Answer:

21. x = 5, 23.

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers