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Lubov Fominskaja [6]
1 year ago
6

Alice has a total of 12 dimes and nickels. She has 2 more nickels than dimes. Which equation represents the given problem situat

ion?
A. c + (c + 2) = 12, where c is the number of dimes

B. c + 2c = 12, where c is the number of nickels

C. c + (c + 2) = 12, where c is the number of nickels

D. c + 2c = 12, where c is the number of dimes
Mathematics
2 answers:
const2013 [10]1 year ago
6 0

Answer:  The correct option is

(A) c+(c+2)=12, where c is the number of dimes.

Step-by-step explanation:  Given that Alice has a total of 12 dimes and nickels and she has 2 more nickels than dimes.

We are to select the correct equation that represents the given problem situation.

Let c represents the number of dimes. Then, the number of nickels will be (c + 2).

Since there are total 12 coins, so the required equation is given by

c+(c+2)=12.

Thus, the required equation is

c+(c+2)=12, where c is the number of dimes.

Option (A) is CORRECT.

Westkost [7]1 year ago
5 0

Answer:

A.

Step-by-step explanation:

She has 2 more nickels then dimes not 2 times more therefore answers B and D are incorrect. C is incorrect because it has that there are 2 more dimes than nickels. A is correct because it says that there are c dimes, and then c +2 nickels.

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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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The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

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°

Since, AM is perpendicular bisector of BC. So,

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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

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