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Ivan
2 years ago
12

I'LL GIVE YOU BRAINIEST! 1 1/5 · x/3 = 9/10 solve for x

Mathematics
2 answers:
joja [24]2 years ago
8 0

\huge\text{Hey there!}

\huge\textbf{Equation:}

\mathbf{1 \dfrac{1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}

\huge\textbf{Solve:}

\mathbf{1 \dfrac{1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}

\mathbf{\rightarrow \dfrac{1\times5+1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}

\mathbf{\rightarrow \dfrac{5 + 1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}

\mathbf{\rightarrow \dfrac{6}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}

\huge\textbf{Simplify it:}

\mathbf{\rightarrow \dfrac{2}{5}x = \dfrac{9}{10}}

\huge\textbf{Multiply \boxed{\bf \dfrac{5}{2}} to both sides:}

\mathbf{\rightarrow \dfrac{5}{2}\times \dfrac{2}{5}x = \dfrac{5}{2}\times \dfrac{9}{10}}

\huge\textbf{Simplify it:}

\mathbf{\rightarrow x = \dfrac{5}{2}\times \dfrac{9}{10}}

\mathbf{\rightarrow x = \dfrac{5\times9}{2\times10}}

\mathbf{\rightarrow x = \dfrac{45}{20}}

\mathbf{\rightarrow x = \dfrac{45\div5}{20\div5}}

\mathbf{\rightarrow x = \dfrac{9}{4}}

\mathbf{x \approx 2 \dfrac{1}{4}}

\huge\textbf{Therefore, your answer should be:}

\huge\boxed{\mathsf{x = }\frak{2 \dfrac{1}{4}}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}

~\frak{Amphitrite1040:)}

shusha [124]2 years ago
7 0

<u>Question</u><u> </u><u>:</u>

  • 1 1/5 . x/3 = 9/10 Solve for x

<u>Solution</u><u> </u><u>:</u>

>> (5 × 1 + 1 / 5) × (x / 3) = 9 / 10

>> (5 + 1 / 5) × (x / 3) = 9 / 10

>> (6 / 5) × (x / 3) = 9 / 10

>> 6 × x / 5 × 3 = 9 / 10

>> 6x / 15 = 9 / 10

>> 2x / 5 = 9 / 10

>> 2x × 10 = 5 × 9

>> 20x = 45

>> x = 45 / 20

>> x = 9 / 4

Therefore,

  • Value of x is 9 / 4 !!
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</span><span>Step 1: First, you must draw an image to help better understand the problem 
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The reason that ship A is moving -35km/h is because ship B acts as an "origin" and as things move closer to the "origin" they become negative and as things move away from the "origin" they become positive.

We know that the rate of Ship A is expressed as <span><span><span>dA</span><span>dt</span></span>=−35</span> and the rate of Ship B is expressed as <span><span><span>dB</span><span>dt</span></span>=25</span>

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Now if we take this <span>−140km</span> and add it to the <span>150km</span> we can see that Ship A is now only <span>10km</span> away from where Ship B began

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This means that Ship B is now <span>100km</span> from where it originally started

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Step 4: Now that we know z, we must differentiate the original equation in order to find the rate at which z in changing

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Now plug in the information

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<span>10<span>(−35)</span>+100<span>(25)</span>=<span>√<span><span>102</span>+<span>1002</span></span></span><span><span>dz</span><span>dt</span></span></span>

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