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3 years ago

Christine drove 260 miles using 12 gallons of gas. At this rate, how many gallons of gas would she need to drive 286 miles?

Mathematics

2 answers:

irinina [24]3 years ago

7 0

Answer:

13.2 gallons

Step-by-step explanation:

286/260 *12 = 13.2

Hatshy [7]3 years ago

6 0

Answer:

about 13.2 gallons

Step-by-step explanation:

Easy way to calculate.

12 gallons divided by 260 miles

Take that answer and multiply it by 286.

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Showing ALL calculations, whether this statement is valid

choli [55]

Answer:

RM79 608.51

Step-by-step explanation:

First year,

RM505 050.00×5%=RM505 050.00×5/100

=RM25 252.50

second year,

(RM505 050.00+RM25 252.50)×5%=RM530 302.50×5/100

=RM26 515.13

third year,

(RM530 302.50+RM26 515.13)×5%=RM556817.63×5/100

=RM27840.88

three years interest,

RM25 252.50+RM26 515.13+RM27840.88=RM79608.51

I don't know that is R or RM 

if it is R ,sorry

Hope can help!

8 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,

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

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

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

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