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Nonamiya
2 days ago
8

If f(x) = 7 + 4x and g (x) = StartFraction 1 Over 2 x EndFraction, what is the value of (StartFraction f Over g EndFraction) (5)

? Eleven-halves StartFraction 27 Over 10 EndFraction 160 270
Mathematics
2 answers:
Svet_ta [12.2K]2 days ago
4 0

Answer:

D) 270

Step-by-step explanation:

congratulations, you're done with this assessment

lawyer [12K]2 days ago
4 0

Response:

\frac{f}{g}(5) = 270 ⇒ Previous solution

Detailed breakdown:

* Given f(x) = 7 + 4x

* Given g(x) = \frac{1}{2x}

* We aim to determine \frac{f}{g}(5)

- Initially, let’s calculate \frac{f}{g}(x)

∵ f(x) = 7 + 4x

∵ g(x) = \frac{1}{2x}

∴ \frac{f}{g}(x)=\frac{7+4x}{\frac{1}{2x}}

- Let’s perform division of the numerator by the denominator

∵ The numerator is 7 + 4x

∵ The denominator is \frac{1}{2x}

∴ (7 + 4x) ÷ \frac{1}{2x}

- Now we will change the division sign to a multiplication sign and take the reciprocal of

the fraction following the division sign

∴ (7 + 4x) × \frac{2x}{1}

∴ \frac{f}{g}(x) = 2x(7 + 4x)

∴ \frac{f}{g}(x) = 14x + 8x²

- Next, substitute x with 5

∴ \frac{f}{g}(5) = 14(5) + 8(5)² = 70 + 200 = 270

∴ \frac{f}{g}(5) = 270

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8 0
9 days ago
If the volume of a cube is increased by a factor of 10, by what factor does the surface area per unit volume change
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Answer:

  (1/10)∛100 ≈ 0.4642

Step-by-step explanation:

For a cubic shape with volume V, the edge length is defined as...

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and the surface area is given by...

  A = 6s² = 6V^(2/3)

The area-to-volume ratio is therefore...

  r1 = A/V = 6V^(2/3)/V = 6V^(-1/3)

When the volume V is increased by a factor of 10, the new area-to-volume ratio becomes...

  r2 = 6(10V)^(-1/3)

Consequently, the change factor for the ratio is...

  r2/r1 = (6(10V)^(-1/3))/(6V^(-1/3)) = 10^(-1/3) = (1/10)∛100

The change in surface area per unit volume results in a factor of (∛100)/10.

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Example

For a cube with side length 2, the corresponding volume equals 2³ = 8, with a surface area of 6·2² = 24. The resulting area-to-volume ratio is 24/8 = 3.

<pif we="" multiply="" the="" edge="" length="" by="" new="" volume="" equals="" and="" surface="" area="" equates="" to="" so="" area-to-volume="" ratio="" becomes...="">

The area-to-volume ratio changed by a factor of (0.3∛100)/3 = (∛100)/10, as previously noted.

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What is the midpoint of a line segment with the endpoints (-6, -3) and (9, -7)?
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. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i
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Answer:

Error made by Andrew: He identified incorrect factors based on the roots.

Step-by-step explanation:

The roots of the polynomial consist of: 3, 2 + 2i, 2 - 2i. By the factor theorem, if a is a root of the polynomial P(x), then (x - a) must be a factor of P(x). According to this premise:

(x - 3), (x - (2 + 2i)), (x - (2 - 2i)) represent the factors of the polynomial.

<pBy simplification, we obtain:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) as the respective factors.

This is where Andrew's mistake occurred. Factors should always be in the form (x - a), not (x + a). Andrew expressed the complex factors incorrectly, resulting in an erroneous conclusion.

Thus, the polynomial can be expressed as:

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7 0
1 month ago
In isosceles △ABC (AC = BC) with base angle 30° CD is a median. How long is the leg of △ABC, if sum of the perimeters of △ACD an
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Important details about isosceles triangle ABC:

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  • In isosceles triangle ABC, the sides AB and BC are equal, meaning AC=BC.
  • The base angles at AB are equal, m∠A=m∠B=30°.

1. Consider the right triangle ACD. The angle adjacent to side AD is 30°, which dictates that the hypotenuse AC is double the length of the opposite side CD relating to angle A.

AC=2CD.

2. Now, for right triangle BCD, the angle next to side BD is also 30°, so hypotenuse BC is twice the opposite leg CD linked to angle B.

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3. To calculate the perimeters of triangles ACD, BCD, and ABC:

P_{ACD}=AC+CD+AD=2CD+CD+AD=3CD+AD;

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4. If the total of the perimeters of triangles ACD and BCD is 20 cm greater than the perimeter of triangle ABC, then

P_{ACD}+P_{BCD}=P_{ABC}+20,\\ \\3CD+AD+3CD+AD=4CD+2AD+20,\\ \\6CD+2AD=4CD+2AD+20,\\ \\2CD=20.

5. Given that AC=BC=2CD, the lengths of legs AC and BC of the isosceles triangles are 20 cm.

Answer: 20 cm.

8 0
29 days ago
Read 2 more answers
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