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ElenaW [278]
1 year ago
11

Solve (5√16)^1 using prime factorisation.​

Mathematics
1 answer:
krok68 [10]1 year ago
7 0

Answer:

2^{\frac{4}{5}}

Step-by-step explanation:

Factor 16 using prime factorisation:

16 ÷ 2 = 8

8 ÷ 2 = 4

4 ÷ 2 = 2

⇒ 16 = 2⁴

Substitute 16  for  2⁴:

\implies (\sqrt[5]{16})^1=(\sqrt[5]{2^4})^1

\textsf{Apply exponent rule}\quad \sqrt[n]{a^b} =a^{\frac{b}{n}}:

\implies (\sqrt[5]{2^4})^1=(2^{\frac{4}{5}})^1

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\implies (2^{\frac{4}{5}})^1=2^{\frac{4}{5}}

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Step-by-step explanation:

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2 years ago
Find the amplitude and period of y = 5 cos (4x)
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Answer: amplitude is 5 and Period is pi/2

Step-by-step explanation:

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2 years ago
Anyone can help me solve this equation using cross multiplying
Natali5045456 [20]

9514 1404 393

Answer:

  x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

  3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}

_____

<em>Comment on "cross multiply"</em>

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "<em>multiply by the product of the denominators</em> and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}

8 0
2 years ago
-3x+10=21 solve for x
Fed [463]

Answer:

Step-by-step explanation:

-3x+10=21

-3x=21-10

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x=11/-3

x=- 11/3

8 0
2 years ago
What is the true solution to the equation below? 2 in e in2×-in e in 10×= in 30 A x=30 B x=75 C x=150 D x=300
hodyreva [135]

Answer:

Option B.

Step-by-step explanation:

Let as consider the given equation:

2\ln e^{\ln 2x}-\ln e^{\ln 10x}=\ln 30

It can be written as

2(\ln 2x)-(\ln 10x)=\ln 30         [\because \ln e^a=a]

\ln (2x)^2-(\ln 10x)=\ln 30        [\because \ln a^b=b\ln a]

\ln \dfrac{4x^2}{10x}=\ln 30        [\because \ln \dfrac{a}{b}=\ln a-\ln b]

\ln \dfrac{2x}{5}=\ln 30

On comparing both sides, we get

\dfrac{2x}{5}=30

Multiply both sides by 5.

2x=150

Divide both sides by 2.

x=75

Therefore, the correct option is B.

7 0
3 years ago
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