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irina [24]
2 years ago
13

A boy band has released two albums. Their first album sold 5.9×10^5 copies. Their second album sold 1.3×10^6 copies. How many to

tal copies has this boy band sold?
Mathematics
2 answers:
PIT_PIT [208]2 years ago
4 0

Answer:

The boy band sold  1.89 ×10^6 copies

Step-by-step explanation:

To get the number of copies we must add 5.9×10^5 copies with 1.3×10^6  copies.  We can´t add them  like that.

To make the addition we have to adjust the power of 10,  so they have the same index

In this case we can change  5.9×10^5 to 0.59×10^6 ( we move the comma to the left , and we add 1 power to the scientific quotation.

So, 0.59×10^6

<u />

<u>+    1.3×10^6  </u>

     1.89 ×10^6

prisoha [69]2 years ago
3 0
The band has sold 1,890,000 copies or 1.89 x 10^6
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Let's call the prices a and c, a for adult...

The cost of an adult ticket is £7 more than the cost of a child ticket:

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86 = 2a + 4(a-7) = 6a - 28

114 = 6a

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Answer: £19

Check:

That implies c=12.

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In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
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Answer:

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Step-by-step explanation:

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The diagram is given below :

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Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

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MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

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sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

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