6 ice cream cones cost $10. How much does 9 cost

On a piece of paper, graph this system of inequalities. Then determine which region contains the solution to the system.

kirill115 [55]

B or C hope its helps...

5 0

1 year ago

There is a point $A$ with positive coordinates such that the sum of the coordinates of $A$ is $14$. If the $x$-coordinate of $A$

ch4aika [34]

Answer:

5

Step-by-step explanation:

Given:

A = (a, 14-a)

P = (3a, a^2 +13a -11)

the slope of AP is 7

a > 0

Find:

a

Solution:

The slope of AP is ...

m = (Py -Ay)/(Px -Ax)

7 = (a^2 +13a -11 -(14 -a))/(3a -a)

14a = a^2 +14a -25

25 = a^2

a = √25 = 5 . . . . . the positive solution

The value of 'a' is 5.

_____

Check

The point A is (a, 14-a) = (5, 9).

The point P is (3a, a^2 +13a -11) = (15, 79)

The slope of AP is (79 -9)/(15 -5) = 70/10 = 7.

6 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

1 year ago

Find the eccentricity, b. identify the conic, c. give an equation of the directrix, and d. sketch the conic.

densk [106]

Answer:

a) 10/3

b) hyperbola

c) x = ± 6/5

Step-by-step explanation:

a) A conic section with a focus at the origin, a directrix of x = ±p where p is a positive real number and positive eccentricity (e) has a polar equation:

$r=\frac{ep}{1\pm e*cos\theta}$

Given the conic equation: $r=\frac{12}{3-10cos\theta}$

We have to make it to be in the form $r=\frac{ep}{1\pm e*cos\theta}$ :

$r=\frac{12}{3-10cos\theta}\\\\multiply\ both\ sides\ by\ \frac{1}{3} \\\\r=\frac{12*\frac{1}{3}}{(3-10cos\theta)*\frac{1}{3}}\\\\r=\frac{12*\frac{1}{3}}{3*\frac{1}{3}-10cos\theta*\frac{1}{3}}\\\\r=\frac{4}{1-\frac{10}{3}cos\theta } \\\\r=\frac{\frac{10}{3}(\frac{6}{5} ) }{1-\frac{10}{3}cos\theta }$

Comparing with $r=\frac{ep}{1\pm e*cos\theta}$

e = 10/3 = 3.3333, p = 6/5

b) since the eccentricity = 3.33 > 1, it is a hyperbola

c) The equation of the directrix is x = ±p = ± 6/5

6 0

1 year ago

The profit earned by a sporting goods outlet is modeled in the graph below, where x is the number of years since the outlet open

LenaWriter [7]

The domain would be x ≥ 0.

This is because the outlet cannot have profit before it was open. Therefore, the growth must be from year 0 to present. If they give a year as starting, you can have an upper limit too, but there is not enough information here to determine that information.

4 0

1 year ago

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