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2 years ago

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Chris is selling chicken sandwiches and hamburgers at the fair in his home town. He has a total of 40 buns so he can sell no mor

e than 40 chicken sandwiches and hamburgers. Each chicken sandwich sells for $4 and each hamburger sells for $2. In order to reach his goal, Chris must make at least $100.

1 answer:

Blizzard [7]2 years ago

7 0

Answer:

x

Step-by-step explanation:

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Thomas graphed the line that represents the equation y=34x.

zloy xaker [14]

Answer:

The ordered pairs represent points on the line are

(4, 3) ⇒ C

(2, $\frac{3}{2}$ ) ⇒ D

(-8, -6) ⇒ E

Step-by-step explanation:

To find the ordered pairs represent points on the line, substitute x by the x-coordinate of each point, if the value of y equals the y-coordinate of the point, then the point is on the line.

∵ The equation is y = $\frac{3}{4}$ x

∵ The ordered pair is (8, $\frac{1}{6}$ )

→ Substitute x by 8

∴ y = $\frac{3}{4}$ (8)

∴ y = 6

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair (8, $\frac{1}{6}$ ) does not represent a point on the line

∵ The ordered pair is ( $\frac{-2}{3}$ , $\frac{1}{2}$ )

→ Substitute x by $\frac{-2}{3}$

∴ y = $\frac{3}{4}$ ( $\frac{-2}{3}$ )

∴ y = $\frac{-1}{2}$

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair ( $\frac{-2}{3}$ , $\frac{1}{2}$ ) does not represent a point on the line

∵ The ordered pair is (4, 3 )

→ Substitute x by 4

∴ y = $\frac{3}{4}$ (4)

∴ y = 3

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (4, 3) represents a point on the line

∵ The ordered pair is (2, $\frac{3}{2}$ )

→ Substitute x by 2

∴ y = $\frac{3}{4}$ (2)

∴ y = $\frac{3}{2}$

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (2, $\frac{3}{2}$ ) represents a point on the line

∵ The ordered pair is (-8, -6 )

→ Substitute x by -8

∴ y = $\frac{3}{4}$ (-8)

∴ y = -6

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (-8, -6) represents a point on the line

5 0

1 year ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$

$A = 2.525 -2.474+9.798 + 2.525 - 2.474$

$A = 9.9\,units^{2}$

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0

2 years ago

Is 16.45 greater than, less than or equal to 16.454

vampirchik [111]

<span>16.45 is less than 16.454. The reason is because 16.454 is 4 thousandths more than 16.45 assuming that both numbers are exact numbers. Although it is only a small amount it still makes 16.45 less than 16.454.</span>

4 0

2 years ago

Read 2 more answers

During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16

jolli1 [7]

To find the time at which both balls are at the same height, set the equations equal to each other then solve for t.
h = -16t^2 + 56t
h = -16t^2 + 156t - 248
-16t^2 + 56t = -16t^2 + 156t - 248
You can cancel out the -16t^2's to get
56t = 156t - 248
=> 0 = 100t - 248
=> 248 = 100t
=> 2.48 = t
Using this time value, plug into either equation to find the height.
h = 16(2.48)^2 + 56(2.48)
Final answer:
h = 40.4736
Hope I helped :)

6 0

2 years ago

Read 2 more answers

Consider the following system of equations: 10 + y = 5x + x2 5x + y = 1 The first equation is an equation of a . The second equa

aleksley [76]

Answer: The first equation is an equation of a parabola. The second equation is an equation of a line.

Explanation:

The first equation is,

$10+y=5x+x^2$

In this equation the degree of y is 1 and the degree of x is 2. The degree of both variables are not same. Since the coefficients of y and higher degree of x is positive, therefore it is a graph of an upward parabola.

The second equation is,

$5x+y=1$

In this equation the degree of x is 1 and the degree of y is 1. The degree of both variables are same. Since both variables have same degree which is 1, therefore it is linear equation and it forms a line.

Therefore, the first equation is an equation of a parabola. The second equation is an equation of a line.

5 0

2 years ago

Read 2 more answers