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rjkz [21]
3 years ago
14

What is 9/45 divided by 7/42

Mathematics
2 answers:
Brrunno [24]3 years ago
8 0
first\ sipmlify\ the\ 
fractions:\\\dfrac{9}{45}=\dfrac{9:9}{45:9}=\dfrac{1}{5}\\\\\dfrac{7}{42}=\dfrac{7:7}{42:7}=\dfrac{1}{6}\\\\\dfrac{9}{45}:\dfrac{7}{42}=\dfrac{1}{5}:\dfrac{1}{6}=\dfrac{1}{5}\cdot\dfrac{6}{1}=\dfrac{1\cdot6}{5\cdot1}=\dfrac{6}{5}=\dfrac{5+1}{5}=\dfrac{5}{5}+\dfrac{1}{5}\\\\=1+\dfrac{1}{5}=\boxed{1\dfrac{1}{5}}
Ghella [55]3 years ago
7 0
6/5. simplify 9/45 to 1/5 and flip 7/42 to 42/7 and multiply 1/5 by 42/7 which would be 42/35 and simplify to 6/5
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ahrayia [7]
So,

#1: n  \geq 3 \frac{11}{16} + 4 \frac{1}{2}

Convert to like improper fractions.
n  \geq  \frac{59}{16} +  \frac{72}{16}

Add.
n  \geq  \frac{131}{16}\ or\ 8 \frac{3}{16}

So, one solution could be 8 \frac{3}{16}.

Another solution could by 9.  There is also 10, 11, 12, etc., and all numbers in between.


#2: k \ \textless \  6  \frac{2}{5} * 15

Convert into improper fraction form.
k \ \textless \ \frac{32}{5} * 15

Multiply.
\frac{(2^5)(3)(5)}{5}

Cross-cancel, and we have our final result.
(2^5)(3) = 96
k < 96

96 is not a solution.

95 is a solution.

So is 94, 93, 92, etc, and all numbers in between.
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3 years ago
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Scorpion4ik [409]
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4 0
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Read 2 more answers
Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer
Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

0=(3x+2)(x-4)

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply (3x+2)(x-4) i.e. 3x^2-12x+2x-8 = 3x^2-10x-8

So, the equation becomes:

0=3x^2-10x-8

Comparing this equation to standard quadratic equation:

ax^2+bx+c=0

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}

The answer is same from both the approaches.

So, the correct answer is:

C) Both

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DIA [1.3K]

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Step-by-step explanation:

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