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Andru [333]
3 years ago
7

Estimate how many times larger 6.1×10^7 is than 2.1×10^−4. PLEASE WILL GIVE BRAINLIEST!!!!!! cmon plsssss

Mathematics
1 answer:
cestrela7 [59]3 years ago
8 0
The answer is 2904. Enjoy
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3 plus 4o hundredths is larger.

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Solve for the variable <br> 3x=-7
Allisa [31]
3x = -7   Divide both sides by 3. 3 goes into seven twice with one-third left                   over. Since it's a negative seven, the answer will also be negative.
x = -2\frac{1}{3}
8 0
3 years ago
Help me please thanks in advance
Xelga [282]

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From up to down ;

_________________________________

The first box :

f(n) = 3. ({2})^{n - 1}

The second box :

f(n) = 2. ({4})^{n - 1}

The third box :

f(n) = 2. ({6})^{n - 1}

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4 0
3 years ago
Which one of these is ordered from greatest to least?
Mama L [17]

Answer:

the answer would be C :)

8 0
4 years ago
Simplify the radical <br> Sqrt 84x^7
pentagon [3]

Answer:

\large\boxed{\sqrt{84x^7}=2x^3\sqrt{21x}}

Step-by-step explanation:

Domain:\ x\geq0\\\\\sqrt{84x^7}=\sqrt{4\cdot21\cdot x^{6+1}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{4\cdot21\cdot x^6\cdot x^1}\\\\\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt4\cdot\sqrt{21}\cdot\sqrt{x^{3\cdot2}}\cdot\sqrt{x}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=2\cdot\sqrt{21}\cdot\sqrt{(x^3)^2}\cdot\sqrt{x}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2\cdot\sqrt{21}\cdot x^3\cdot\sqrt{x}=2x^3\sqrt{21x}

6 0
3 years ago
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