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seropon [69]
3 years ago
6

25 POINTS AND BRAINLIEST! PLEASE HELP ME !!! answer all questions please !

Mathematics
2 answers:
dlinn [17]3 years ago
8 0

Answer:

One inch represented 3 feet in the first scale, but now 1 inch now represents 4 feet in the second scale.

Step-by-step explanation:

Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

PART A : This formula shows how to find the actual length of the microchip:

actual length = scale length × reciprocal of scale factor.

The scale is 2 inches to 1 centimeter, so the scale factor is . 2in/1cm  The reciprocal of the scale factor is . 1cm/2in The length of the chip in the scale drawing is 6 inches.

To find the actual length of the microchip, substitute the values into the formula:

actual length  =  scale length × reciprocal of scale factor

= 6 in.  x  (1cm/2in)

= 3 cm

The actual length of the microchip is 3 centimeters.

PART B:

The scale is 2 inches to 1 centimeter, so the scale factor is .2in/1cm The reciprocal of the scale factor is .1cm/2in The width of the chip in the scale drawing is 4 inches.

To find the actual width of the microchip, substitute the values into the formula:

actual width  = scale width × reciprocal of scale factor

= 4 in. x  (1cm/2in)

= 2 cm

The actual width of the microchip is 2 centimeters.

PART C:

The new scale is 3 inches to 1 centimeter, so the new scale factor is . 3in/ 1cm The actual microchip is 3 centimeters long. Substitute these values into the following equation to find the scale length:

scale length = actual length × scale factor

= 3 cm  x (3in/1cm)

= 9 in.

The length of the microchip in the new scale drawing is 9 inches.

PART D:

The width of the microchip in the new scale drawing is 6 inches.

PART E:

In the original scale drawing, the microchip was 6 inches long. In the new scale drawing, the microchip is 9 inches long. Set up the ratio:

ratio .  6in/9 in 3in/2in

In the original scale drawing, the microchip was 4 inches wide. In the new scale drawing, the microchip is 6 inches wide. Set up the ratio:

ratio .  4in/6in  2in/3in

The ratio of the original scale drawing measurements to the new scale drawing measurements is 2 inches to 3 inches.

PART F:

If a scaled measurement is greater than the actual measurement it represents, then the scale drawing is larger than the actual object.

If a scaled measurement is less than the actual measurement it represents, then the scale drawing is smaller than the actual object.

PLEASE GIVE ME BRAINLIEST :))

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How do you do this question?
Lena [83]

Step-by-step explanation:

The Taylor series expansion is:

Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!

f(x) = 1/x, a = 4, and n = 3.

First, find the derivatives.

f⁽⁰⁾(4) = 1/4

f⁽¹⁾(4) = -1/(4)² = -1/16

f⁽²⁾(4) = 2/(4)³ = 1/32

f⁽³⁾(4) = -6/(4)⁴ = -3/128

Therefore:

T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!

T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³

f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0.  So we can eliminate the top left option.  That leaves the other three options, where f(x) is the blue line.

Now we have to determine which green line is T₃(x).  The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).

The bottom right graph is the only correct option.

3 0
3 years ago
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Step-by-step explanation: I think

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Step-by-step explanation:


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We have
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