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irina1246 [14]
3 years ago
13

Simplify: Cube root of (128a^13b^6)

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
7 0
\sqrt[3]{128a^{13}b^{6}} \\\sqrt[3]{64a^{12}b^{6} * 2a} \\\sqrt[3]{64a^{12}b^{6}}\sqrt[3]{2a} \\4a^{4}b^{2}\sqrt[3]{2a}
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A hot tub is draining, and the water level is decreasing at a constant rate. After 9.3 minutes, the tub contains 401.92 gallons.
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Step-by-step explanation:

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3 years ago
There are 6 gallons in a tank that is 20% full. How many gallons are in a full tank? Use a double number line to show your work.
Alex17521 [72]

Answer: 30 gallons

Step-by-step explanation:

Let the number of gallons that are in a full tank be represented by x.

Since there are 6 gallons in a tank that is 20% full, this will be set up in an equation as:

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3 years ago
Please can you help me with this
Orlov [11]
4a-87=a

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5 0
3 years ago
Anyone knows how to do questions 7 and 8? 15 pts!!
Oksi-84 [34.3K]
7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
3 0
3 years ago
Read 2 more answers
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