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3 years ago

The sun is about 93 x 10^6 miles from earth. what is this distance written as a whole number?

2 answers:

5 0

<span>This distance would be written as a whole number as 93,000,000. To get this number, you would use the first two numbers and then add 6 zeros to the end to account for the 10^6. 10^6 is equal to 10 multiplied by 10 six times with each product being multiplied by 10. This can also be displayed as 1E6. This number is then multiplied by 93 to get 93,000,000.</span>

agasfer [191]3 years ago

3 0

For this case we have the number in exponential notation:

$93 * 10 ^ 6$

We observe that the exponent of the exponential notation is positive.

Therefore, to express the number as a whole number, we must move the comma 6 spaces to the right.

We have then:

$93 * 10 ^ 6 = 93,000,000$

Answer:

this distance written as a whole number is:

$93 * 10 ^ 6 = 93,000,000$ miles

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3 years ago

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vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

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3 years ago

Doug’s family buys 7 postcards while on vacation. Each postcard costs 0.25$ including tax. How can Doug use the number line to f

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Answer: Total cost is $1.75

Explanation:

Since we have given that

Number of postcards bought by Doug's family= 7

Cost of each postcard including tax = $0.25

Total cost is given by

$7\times 0.25=\$1.75$

Now, we want to show this in number line :

On the number line put all the number with a difference of 0.25 and run the number line seven times and we get the answer.

Each step is of length 0.25 and there are 7 jumps to reach required answer.

So, after 7 jumps we reach at 1.75 which is the required answer.

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il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

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$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

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3 years ago