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kirill [66]
2 years ago
4

Raymond wanted to buy 8 t-shirt but he was short of $8.10. instead he bought 5 T-shirt and had $12.60 left. how much would he ne

ed to pay for 20 t-shirt?
Mathematics
1 answer:
BigorU [14]2 years ago
5 0

Answer:

$138

Step-by-step explanation:

Let the cost if a tee shirt be $T.

Since the amount of money Raymond have is constant,

8T -8.10= 5T +12.60

8T -5T= 12.60 +8.10

3T= 20.7 <em>(</em><em>simplify</em><em>)</em>

T= 20.7 ÷3 <em>(</em><em>÷</em><em>3</em><em> </em><em>throughout</em><em>)</em>

T= 6.90

Thus, cost of one t-shirt is $6.90.

Cost of 20 t-shirts

= 20(6.90)

= $138

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Answer:

<u>The number of intersections = ∞</u>

Step-by-step explanation:

Given the system of equations:

0.5 x + 5y = 6

3x + 30y = 36

By multiplying the first equation by 6

∴ 6 (0.5 x + 5y) = 6 * 6

∴ 3x + 30y = 36

So, Those equations are "Dependent", because they are really the same equation, just multiplied the first equation by 6.

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<u>See the attached figure.</u>

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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:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

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

Since, AM is perpendicular bisector of BC. So,

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}

Now, in ΔBMP :

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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