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Sergio039
2 months ago
5

Conner and Jana are multiplying (3568)(39610). Conner's Work Jana's Work (3568)(39610) = 35 + 968 + 10 = 314618 (3568)(39610) =

35⋅968⋅10 = 345680 Is either of them correct? Explain your reasoning.
Mathematics
2 answers:
AnnZ [12.3K]2 months ago
8 0

Answer:

Neither Conner nor Jana provided the correct solution.

Step-by-step explanation:

Conner’s error lies in adding exponents, which only applies when multiplying powers with the same base. Here, the bases are different, so this rule doesn’t apply. Jana’s mistake involves incorrect exponent multiplication; she multiplied 5 by 9 and 8 by 10 mistakenly, instead of correctly multiplying 5 by 8 and 9 by 10.

My calculated answers are 6^40 and 6^90 respectively.

(3^5 × 6^8) = 3^5 × 6^8 = 6^40

(3^9 × 6^10) = 3^9 × 6^10 = 6^90

This problem was quite challenging, but I hope this explanation helps, even if the question is older.

lawyer [12.5K]2 months ago
4 0

Answer:

Step-by-step explanation:

Neither Conner nor Jana correctly broke down the numbers for multiplication. The correct product of 3568 times 39610 is 141,328,480.

Both attempted to use the partition method, but failed to divide the numbers properly.

Specifically, 3568 should be split as 3000 + 500 + 60 + 8,

and 39610 as 30000 + 9000 + 600 + 10.

Then, multiply each part of 3568 by each part of 39610, such as:

3000 × 30000

3000 × 9000

3000 × 600

3000 × 10

and continuing with all combinations.

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A piece of climbing equipment at a playground is 6 feet high and extends 4 feet horizontally. A piece of climbing equipment at a
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Response: The following comparison is made.

Detailed explanation:

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In contrast, a climbing structure in the gym stands 10 feet tall, extending 6 feet horizontally.

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Answer: Option C - Construction Y, as point E serves as the circumcenter of triangle LMN.

Point E is ideal for the warehouse since it maintains equal distances from the three stores located at L, M, and N.

Step-by-step explanation:

To effectively tackle an algebra problem, establishing a geometric illustration frequently proves beneficial. Geometry states that three points define a circle; meaning, there is one unique circle encompassing all three non-collinear points. Identifying the point equidistant from these three points corresponds to finding the circle's center that passes through them (as all points on the circle share equal distance from the center).

Considering our points L, M, and N, draw the segments LM, LN, and MN to outline a triangle. Construct the perpendicular bisectors of any two line segments, and their intersection, point E, will indicate the center of this circle.

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20 days ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
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Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The equation of the line with the LARGEST slope that is tangent to both graphs is

(b) The equation of the second line tangent to both graphs is:

Solution:

- First, let's calculate the derivatives for the two functions provided:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Given that the derivatives of both functions are dependent on the x-value, we will pick a common point x_o for both f(x) and g(x). This point is ( x_o, g(x_o)). Thus,

                                g'(x_o) = -2*(x_o - 2)

- Next, we will determine the slope of a line that is tangent to both graphs at point (x_o, g(x_o) ) on g(x) and at ( x, f(x) ) on f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now we need to set the slope from our equation equal to the derivatives we calculated earlier for each function:

                                m = f'(x) = g'(x_o)

- We will work through the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now we can subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Rearranging gives us:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o,     x_o = 10x + 2    

- For x_o = 10x + 2,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Now solving the above quadratic equation:

                                 x = -0.0574, -0.387      

- The maximum slope occurs at x = -0.387, with the line’s equation being:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- The second tangent line is:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

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