Find the slope of the line that passes through the points (2,4) and (6,+2)

A candy store makes a 9-pound mixture of gummy candy, jelly beans, and hard candy. The cost of gummy candy is $2.00 per pound, j

Musya8 [376]

Let

x--------> the amount of gummy candy in pounds

y--------> the amount of jelly beans in pounds

z--------> the amount of hard candy in pounds

we know that

$x+y+z=9$ --------> equation $1$

$2x+3y+3z=23$ --------> equation $2$

$x=2y$ --------> equation $3$

Substitute equation $3$ in equation $1$ and equation $2$

$[2y]+y+z=9$ -----> $3y+z=9$ ------> equation $4$

$2[2y]+3y+3z=23$ -----> $7y+3z=23$ ------> equation $5$

using a graphing tool ------> Solve the system of equations

see the attached figure

the solution is the point $(3,2)$

$z=3\ pounds\\y=2\ pounds$

<u>Find the value of x</u>

$x=2y$

$x=2*2=4\ pounds$

therefore

<u>the answer is</u>

the amount of gummy candy is $4\ pounds$

the amount of jelly beans is $2\ pounds$

the amount of hard candy is $3\ pounds$

8 0

3 years ago

Read 2 more answers

Can someone help me with this problem

Ede4ka [16]

-13/24 because u have to multiply to get the same denominator which would end up being -15/24 and 2/24 add that to get -13/24

7 0

3 years ago

What is the LCM of 4,8,14?

Tamiku [17]

The LCM of 4,8,14 is 56

M of 4 are 4,8,12,16,20,24,28,32,36,40,44,48,52,56
M of 8 are 8,16,32,48,56,64
M of 14 are 14,28,42,56

I hope this helps you

5 0

3 years ago

Simplify 2x^2-1+4x^2-5

alekssr [168]

Answer:

6x^2-6

Step-by-step explanation: algebra

8 0

3 years ago

Please help w this! Its a calculus question! look at the picture for the problem,

neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18$

Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is

$\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18$

so the total area is, once again, $36\pi-72$ .

An even simpler way is to subtract the area of the square from the area of the circle.

$\pi6^2-(6\sqrt2)^2=36\pi-72$

6 0

3 years ago

Find the slope of the line that passes through the points (2,4) and (6,+2)​

Find the slope of the line that passes through the points (2,4) and (6,+2)