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3 years ago

A bag contains 12 red marbles the ratio of red marbles to blue marbles is 4:3 find the number of blue marbles in the bag

Mathematics

2 answers:

skelet666 [1.2K]3 years ago

8 0

Answer:

4 blue marbles

Step-by-step explanation:

4:3

4=red marbles

3=blue marbles

total=12

12/3 to get the number of blue marbles

yKpoI14uk [10]3 years ago

4 0

9 blue marbles are in the bag

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Dennis_Churaev [7]

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Step-by-step explanation:

Let the number of months equalising both plans fee = x

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4 years ago

A supplementary angle has three angles one being 44.4 another being 7x and another is 8x. I'm having a hard time.

natima [27]

Answer:

x= 9.04°

7x = 63.28°

8x = 72.32

44.4° + 63.28° + 72.32° = 180°

Step-by-step explanation:

The sum of supplementary angles is 180°

Set up an equation with the given information, then solve for x.

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3x%5E%7B8%7D%2A7x%5E%7B3%7D%20%7D%7B3x%5E%7B6%7D%2A7%20%7D" id="TexFormula1" title="

denis23 [38]

Answer:

$a = \dfrac{x^2}{3} \quad \textsf{and} \quad b = \dfrac{x^3}{7}$

Step-by-step explanation:

Given expression:

$\dfrac{3x^{8} \cdot 7x^{3} }{3x^{6} \cdot 7 }$

There are several ways this problem can be approached, and therefore many different answers. The goal is to reduce the given expression to a simple product of two terms in x, set that to the given form 3a⋅7b then solve for a and b.

$\implies \dfrac{3x^{8} \cdot 7x^{3} }{3x^{6} \cdot 7}$

Cancel the common factors 3 and 7:

$\implies \dfrac{\diagup\!\!\!\!3x^{8} \cdot \diagup\!\!\!\!7x^{3} }{\diagup\!\!\!\!3x^{6} \cdot \diagup\!\!\!\!7}$

$\implies \dfrac{x^8 \cdot x^3}{x^6}$

Separate the fraction:

$\implies \dfrac{x^8}{x^6} \cdot x^3$

$\textsf{Apply the quotient rule of exponents} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies x^{8-6} \cdot x^3$

$\implies x^{2} \cdot x^3$

Now equate the simplified expression to the given form:

$\implies x^{2} \cdot x^3=3a \cdot 7b$

Therefore:

$\begin{aligned}x^{2} &=3a \:\: &\textsf{ and }\:\: \quad x^3 &=7b\\ \Rightarrow a & = \dfrac{x^2}{3} & \Rightarrow b & = \dfrac{x^3}{7}\end{aligned}$

However, we could also separate them as:

$\begin{aligned}x^{3} &=3a \:\: &\textsf{ and }\:\: \quad x^2 &=7b\\ \Rightarrow a & = \dfrac{x^3}{3} & \Rightarrow b & = \dfrac{x^2}{7}\end{aligned}$

Another way of writing them would be to go back a few steps and separate the fraction in x terms differently:

$\implies \dfrac{x^8 \cdot x^3}{x^6}=x^8 \cdot \dfrac{x^3}{x^6}=x^8 \cdot x^{-3}$

Therefore, this would give us:

$\begin{aligned}x^{8} &=3a \:\: &\textsf{ and }\:\: \quad x^{-3} &=7b\\ \Rightarrow a & = \dfrac{x^8}{3} & \Rightarrow b & = \dfrac{1}{7x^3}\end{aligned}$

As the given expression reduces to x⁵, we can separate the x term in any way we like, so long as the coefficient of a is ¹/₃ and the coefficient of b is ¹/₇. Therefore, there are many possible answers.

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