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2 years ago

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The number of pens sold at the school bookstore on Monday, p, increased 45% on Tuesday. The expression p+0.45p represents the nu

mber of pens sold on Tuesday. What is another expression that can be used to represent the number of pens sold on Tuesday?

1 answer:

butalik [34]2 years ago

8 0

Answer:

p + (45/100) p or 1.45 p

Step-by-step explanation:

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taylor bought 2 bottles of orange juice and a bottle of apple juice for $6.55. The bottle of apple juice cost $0.35 less than th

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$6.90 :)

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At a real estate agency, an agent sold a house for $323,000. The commission rate is 6.5% for the real estate agency and the c

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Step-by-step explanation:

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3 years ago

In 2012 your car was worth $10,000. In 2014 your car was worth $8,850. Suppose the value of your car decreased at a constant rat

monitta

Answer:

The function to determine the value of your car (in dollars) in terms of the number of years t since 2012 is:

$f(t) = 10000(0.9407)^t$

Step-by-step explanation:

Value of the car:

Constant rate of change, so the value of the car in t years after 2012 is given by:

$f(t) = f(0)(1-r)^t$

In which f(0) is the initial value and r is the decay rate, as a decimal.

In 2012 your car was worth $10,000.

This means that $f(0) = 10000$ , thus:

$f(t) = 10000(1-r)^t$

2014 your car was worth $8,850.

2014 - 2012 = 2, so:

$f(2) = 8850$

We use this to find 1 - r.

$f(t) = 10000(1-r)^t$

$8850 = 10000(1-r)^2$

$(1-r)^2 = \frac{8850}{10000}$

$(1-r)^2 = 0.885$

$\sqrt{(1-r)^2} = \sqrt{0.885}$

$1 - r = 0.9407$

Thus

$f(t) = 10000(1-r)^t$

$f(t) = 10000(0.9407)^t$

7 0

3 years ago

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

IRISSAK [1]

<u>Prove that:</u>

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

<u>Proof: </u>

We know that, by Law of Cosines,

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

<u>Taking</u><u> </u><u>LHS</u>

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

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

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

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

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

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

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

<em>On combining the fractions,</em>

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

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

<em>Regrouping the terms,</em>

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

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

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

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago