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7nadin3 [17]
2 years ago
12

Which expression is equivalent to x Superscript negative five-thirds? StartFraction 1 Over RootIndex 5 StartRoot x cubed EndRoot

EndFraction StartFraction 1 Over RootIndex 3 StartRoot x Superscript 5 Baseline EndRoot EndFraction Negative RootIndex 3 StartRoot x Superscript 5 Baseline EndRoot Negative RootIndex 5 StartRoot x cubed EndRoot
Mathematics
2 answers:
Soloha48 [4]2 years ago
8 0

Answer:

B

Step-by-step explanation:

Anastasy [175]2 years ago
4 0

Option B : \frac{1}{\sqrt[3]{x^{5} } } is the expression equivalent to x^{-\frac{5}{3}

Explanation:

The given expression is x^{-\frac{5}{3}

Rewriting the expression x^{-\frac{5}{3} using the exponent rule, $a^{-b}=\frac{1}{a^{b}}$

Hence, we get,

\frac{1}{x^{\frac{5}{3} } }

Simplifying, we get,

\frac{1}{\left(x^{5}\right)^{\frac{1}{3}}}

Applying the rule, a^{\frac{1}{n}}=\sqrt[n]{a}

Thus, we have,

\frac{1}{\sqrt[3]{x^{5} } }

Now, we shall determine from the options that which expression is equivalent to x^{-\frac{5}{3}

Option A: \frac{1}{\sqrt[5]{x^{3} } }

The expression \frac{1}{\sqrt[5]{x^{3} } } is not equivalent to simplified expression  \frac{1}{\sqrt[3]{x^{5} } }

Thus, the expression \frac{1}{\sqrt[5]{x^{3} } } is not equivalent to x^{-\frac{5}{3}

Hence, Option A is not the correct answer.

Option B: \frac{1}{\sqrt[3]{x^{5} } }

The expression \frac{1}{\sqrt[3]{x^{5} } } is equivalent to the simplified expression  \frac{1}{\sqrt[3]{x^{5} } }

Thus, the expression \frac{1}{\sqrt[3]{x^{5} } } is equivalent to x^{-\frac{5}{3}

Hence, Option B is the correct answer.

Option C: -\sqrt[3]{x^5}

The expression -\sqrt[3]{x^5} is not equivalent to the simplified expression \frac{1}{\sqrt[3]{x^{5} } }

Thus, the expression -\sqrt[3]{x^5} is not equivalent to x^{-\frac{5}{3}

Hence, Option C is not the correct answer.

Option D: -\sqrt[5]{x^3}

The expression -\sqrt[5]{x^3} is not equivalent to the simplified expression \frac{1}{\sqrt[3]{x^{5} } }

Thus, the expression -\sqrt[5]{x^3} is not equivalent to x^{-\frac{5}{3}

Hence, Option D is not the correct answer.

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What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve.
zvonat [6]

Answer:

x=-2.38

x=-4.62


Step-by-step explanation:

The question is  (2x+3)^2+8(2x+3)+11=0

<em>We let u=2x+3, so the equation becomes:</em>

u^2+8u+11=0

Where a=1, b=8, c=11


Putting it in the quadratic formula, we have:

<u>Quadratic formula:</u> \frac{-b+-\sqrt{b^2-4ac} }{2a}

Substituting we have: \frac{-8+-\sqrt{(8)^2-4(1)(11)} }{2(1)}\\=\frac{-8+-\sqrt{20} }{2}\\=\frac{-8+-2\sqrt{5} }{2}\\=-4+\sqrt{5}, -4-\sqrt{5} }


<em>We let u=2x+3, so x is:</em>

<em>u=2x+3\\(-4+\sqrt{5})=2x+3\\x=\frac{-7+\sqrt{5}}{2}=-2.38</em>

<em>and</em>

<em>u=2x+3\\(-4-\sqrt{5})=2x+3\\x=\frac{-7-\sqrt{5}}{2}=-4.62</em>


The solutions of the equation is x=-2.38 (rounded to 2 decimal places), and x=-4.62 (rounded to 2 decimal places)

6 0
2 years ago
PQ= 2x +1 and QR= 5x - 44; find PQ
Lena [83]

2x+1=5x-44 that would be your equation

next start to simplify 

subtract 1 from both sides so you are left with 2x=5x-45

then subtract 5x from both sides and you have -3x=-45

then finally divide -3 from -45 to get x=15

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2 sandwiches and 3 croissants costs £11.
valentina_108 [34]

Answer:

The price of a sandwich is £2.5 and the price of a croissant is £2

Step-by-step explanation:

Let s be the number of sandwiches and c be the number of croissants

Then according to given statements the equations will be:

2s+3c = 11\ \ \ \ \ \ \ Eqn\ 1\\3s+6c =19.50\ \ \ Eqn\ 2

Multiplying equation 1 by 2 and subtracting equation 2 from the new equation

2(2s+3c) = 11*2\\4s+6c = 22

Now

4s+6c-(3s+6c) = 22-19.50\\4s+6c-3s-6c = 2.5\\s = 2.5

Putting s=2.5 in equation 1

2(2.5)+3c = 11\\5+3c = 11\\3c = 11-5\\3c = 6\\\frac{3c}{3} = \frac{6}{3}\\c = 2

Hence,

The price of a sandwich is £2.5 and the price of a croissant is £2

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Mrs. Anderson Surveyed her class and asked each student. "How many hours do you spend
inysia [295]

Answer:

Scatter plot graph, both 8 and 12 seem less likely to fit they are longer hours in the time axis x, but if we draw a straight line through 5hrs with a ruler, we see a mirror image of 4 to 6 and see less activity in larger amounts of higher hours spent between 7-12

Step-by-step explanation:

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Solve y = 7x-3w for x
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Y=7x-3w
y+3w=7x
y+3w/7=x
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