A bookstore keeps track of books sold in one day? Approximately how many books sold were hardcover nonfiction?

Answers

Answer 1

Answer:

Answer:

27%

Step-by-step explanation:

The total books sold were 182 and the amount of books sold that were nonfiction hardcover were 49 as represented in the frequency table.

To find the percentage we must divide the nonfiction hardcover by the total books.

49/182≈0.269 or 0.27

Multiply the decimal by a 100 to change it into a percentage, so our answer is 27%.

Answer 2

Answer:

Answer:a 27%

Step-by-step explanation:

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Which graphs represent functions?

Answers

i would say the answer is c. only graph d ,, think of the line test to see if a graph is a function or not

Final answer:

Graphs that represent functions have one input corresponding to one output. Examples include straight lines, parabolas, and sine waves.

Explanation:

Graphs that represent functions are those in which every input has exactly one output. In other words, there can only be one value of y for each value of x. For example, a straight line, a parabola, or a sine wave are graphs that represent functions.

On the other hand, graphs that do not represent functions may have one input value mapping to multiple output values or no output values at all. Examples of such graphs include circles, ellipses, or a graph with one vertical line intersecting it at multiple points.

It's important to note that in a function, the vertical line test can be used to determine if a graph represents a function. If any vertical line intersects the graph at more than one point, then the graph does not represent a function.

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QuestionGiven that tan(0) =5/12
and 0 is in Quadrant III. what is cos(0)? Write your answer in exact form. Do not round.
Provide your answer below:

Answers

Answer:

cosΘ = - $(12)/(13)$

Step-by-step explanation:

Given that Θ is in the third quadrant then cosΘ < 0

Given

tanΘ = $(5)/(12)$ = $(opposite)/(adjacent)$

Then 5 and 12 are the legs of a right triangle (5- 12- 13 ) with hypotenuse = 13

Thus

cosΘ = - $(adjacent)/(hypotenuse)$ = - $(12)/(13)$

Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9) The amount of work required is ____ (Simplify your answer.)

Answers

Answer:

Step-by-step explanation:

We want to compute the curve integral (or line integral)

$\bf \int_(C)F$

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write

C = $\bf C_1+C_2$

where

$\bf C_1$ = line segment from (1, 5) to (0, 0)

$\bf C_2$ = line segment from (0, 0) to (0, 9)

so,

$\bf \int_(C)F=\int_(C_1)F+\int_(C_2)F$

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1

Hence $\bf C_1$ can be parameterized as

$\bf r_1(t) = (1-t, 5-5t)$ for 0 ≤ t ≤ 1

and $\bf C_2$ can be parameterized as

$\bf r_2(t) = (0, 9t)$ for 0 ≤ t ≤ 1

The derivatives are

$\bf r_1'(t) = (-1, -5)$

$\bf r_2'(t) = (0, 9)$

and

$\bf \int_(C_1)F=\int_(0)^(1)F(r_1(t))\circ r_1'(t)dt=\int_(0)^(1)(5-5t,t-1)\circ (-1,-5)dt=0$

$\bf \int_(C_2)F=\int_(0)^(1)F(r_2(t))\circ r_2'(t)dt=\int_(0)^(1)(9t,0)\circ (0,-9)dt=0$

In consequence,

$\bf \int_(C)F=0$

Final answer:

The work done by the force field F = (y, -x) along the given path is -5 Joules. This was calculated by breaking the path into two segments and calculating the work done for each segment.

Explanation:

To calculate the work done by the force field F = (y, -x) when moving an object along a specific path, we utilize the concept of the line integral or the dot product of the force and the displacement vector. We can break down the given path into two line segments and solve each separately.

The first segment is from (1, 5) to (0, 0). Only the x component of the displacement is non-zero here, the force is F = (5, -1). Thus the work done on this segment is given by W = F.d = Fd cos θ = -(5 N)(1 m)(cos(180)) = -5 J, where J stands for Joules, the unit of work or energy.

The second segment is from (0, 0) to (0, 9). The force and displacement are perpendicular so the work done is 0.

By adding the work done on these two segments, we arrive at the total work done: W_total = -5 J + 0 J = -5 J

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A 15° sector in a circle has an area of 11.9 m2. What is the area of the circle?

Answers

A 15° sector in a circle has an area of 11.9 m2. What is the area of the circle?
The area of the circle will be found as follows:
the fraction of the sector=15/360=1/24
the area of the circle will be:
area=(fraction of the circle)/(fraction of the sector)*(area of the sector)
=1/(1/24)×11.9
=285.6 m²

Reynaldo rode his bike 2 miles north and 3 miles east. Which equation should he use to find the distance, d, that takes himdirectly back home?
2²+3² - of
3²-2²-2
of +2² - 3²
2 + 3²= 2²
Answer