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PilotLPTM [1.2K]

2 years ago

14

Lisa has a scholarship that pays 75% of her tuition for all four years she attends college. What is the total amount the scholar

ship is worth if Lisa's classes cost 12,000 per year

1 answer:

12345 [234]2 years ago

3 0

Since you're given the total cost per year (12,000) you multiply it by 4 so that you get the total amount for the four years. e.g. (4 x 12,000) = 480,000
Now you calculate the 75% from this amount because we want to obtain the amount of money the scholarship is worth. That is 480,000 x 75% = 360,000
or 480,000 x 0.75 = 360,000
So, the scholarship is of 360,000

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50 POINT QUESTION

LenKa [72]

Answer:

0.04 or 4%

Step-by-step explanation:

point estimate of proportion:

(0.6+0.68)/2

1.28/2

0.64

Margin of error:

0.68 - 0.64 = 0.04

0.04×100 = 4%

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

Two automobiles leave the same city simultaneously and both head towards another city. The speed of one is 10 km/hour greater th

Veronika [31]

Answer:

5091 Km/hr and 505 km/hr

Step-by-step explanation:

Speed = Distance / Time

Let the speed of first automobile be 'x' and that of the second be 'y'

Since speed of one is 10 times greater than the other. therefore;

⇒ x = 10 y

also let time for faster automobile be 'T' and time for slower auto mobile be 't'

Since first arrive one hour earlier than second, therefore;

⇒ t = T + 1

⇒ For first automobile; $x X T = 560$ ; substituting for 'x' and 'T'. Therefore;

⇒ $10y (t+1) = 560$

⇒ For Second automobile; $y X t = 560$

⇒ $y = \frac{560}{t}$

⇒ $10(\frac{560}{t})(t + 1) = 560$

⇒ 5600 + $\frac{5600}{t}$ = 560

⇒ 5600 - 560 = - $\frac{5600}{t}$

⇒ t = 1.11 hr

also ; T = 1.11 - 1 = 0.11 hr

Speed of 1st auto = 560/0.11 = 5091 km /hr

Speed of 2nd auto = 560/1.11 = 505 km/hr

8 0

2 years ago

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7 0

3 years ago

How much is 1/11 of 55

shepuryov [24]

Answer:5

Step-by-step explanation:

you just divide 55 by 11

8 0

2 years ago

Read 2 more answers