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inysia [295]
3 years ago
12

translate this sentence into an inequality. A cheetah can reach a speed of 70 mph. however, this speed can be maintained for no

more than 1,640 feet. A. d>1,640 B. d_<1,640 C. d<1,640 D. d_>1,640
Mathematics
1 answer:
Gre4nikov [31]3 years ago
7 0
<h2>Hello!</h2>

The answer is:

B. d\leq 1,640ft

<h2>Why?</h2>

From the statement we know that the cheetah can reach a speed of 70 mph, but it can be maintained for no more than 1,640 feet.

The expression "no more than" means that at least it can be reached but never exceeded, it involves that the distance can be less or equal than 1,640 feet but never more than that.

So, the correct option is:

B. d\leq 1,640ft

Have a nice day!

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Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

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2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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