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Rzqust [24]
3 years ago
11

There are 15 copies of a popular cd left to be sold in the store this is between 1% and 1.5% of the original number of copies of

the CD in the store the original number of CDs with between two numbers
Mathematics
1 answer:
Triss [41]3 years ago
3 0

Answer: The original number of CDs is between 1000 and 1500.

Step-by-step explanation:

Since we have given that

Number of copies of a popular CD left to be sold in the store = 15

According to question, this is between 15 and 1.5% of the original number of copies of the CD in the store.

Let the original number of copies of the CD in the store be 'x'.

Now, 15 is equal to 1% of the original number of copies of the CD, it becomes

1\%\ of\ x=15\\\\\frac{1}{100}\times x=15\\\\x=15\times 100\\\\x=1500

Or if 15 is equal to 1.5% of the original number of copies of the CD, it becomes

1.5\%\ of\ x=15\\\\\frac{1.5}{100}\times x=15\\\\\frac{15}{1000}x=15\\\\x=\frac{15\times 1000}{15}\\\\x=1000

Hence, the original number of CDs is between 1000 and 1500.

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klasskru [66]

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Triangle ABC has vertices at A(-3,4)B(4,-2)C(8,3).The triangle is translated 4 units down and 3 units left . Which rule represen
FinnZ [79.3K]

Answer:

see explanation

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Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
3 years ago
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