The question.

Your computer-supply store sells two types of inkjet printers. The first, type A, costs $237 and you make a $22 profit on each o

Vera_Pavlovna [14]

The minimum cost option can be obtained simply by multiplying the number of ordered printers by the cost of one printer and adding the costs of both types of printers. Considering the options:
69 x 237 + 51 x 122 = 22,575
40 x 237 + 80 x 122 = 19,240
51 x 237 + 69 x 122 = 20,505
80 x 237 + 40 x 122 = 23,840
Therefore, the lowest cost option is to buy 40 of printer A and 80 of printer B

The equation, x + 2y ≤ 1600 is satisfied only by options:
x = 400; y = 600
x = 1600
Substituting these into the profit equation:
14(400) + 22(600) - 900 = 17,900
14(1600) + 22(0) - 900 = 21,500
Therefore, the option (1,600 , 0) will produce greatest profit.

7 0

3 years ago

Slope 5/2; (-6,7), what is the answer?

nirvana33 [79]

Answer: y = 5/2 x + 22

Step-by-step explanation: Use the formula y=mx+b in order to solve for b. Then plug in the known values.

Graph down below.

I hope this help you out. If not, I am so sorry.

7 0

3 years ago

If the distance from A (5,6) to B (1, b) is twice the distance from B to

Nimfa-mama [501]

Answer:

The possible values of $b$ are -2.944 and -9.055, respectively.

Step-by-step explanation:

From statement we know that $AB = 2\cdot BC$ . By Analytical Geometry, we use the equation of a line segment, which is an application of the Pythagorean Theorem:

$AB = 2\cdot BC$

$\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}} = 2\cdot \sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}$ (1)

Where:

$x_{A}$ , $x_{B}$ , $x_{C}$ - x-Coordinates of points A, B and C.

$y_{A}, y_{B}, y_{C}$ - y-Coordinates of points A, B and C.

$(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2} = 4\cdot (x_{C}-x_{B})^{2}+4\cdot (y_{C}-y_{B})^{2}$

Then, we expand and simplify the expression above:

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} +x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot (x_{C}^{2}-2\cdot x_{C}\cdot x_{B}+x_{B}^{2})+4\cdot (y_{C}^{2}-2\cdot y_{C}\cdot y_{B}+y_{B}^{2})$

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} + x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot x_{A}^{2}-8\cdot x_{C}\cdot x_{B}+4\cdot x_{B}^{2}+4\cdot y_{C}^{2}-8\cdot y_{C}\cdot y_{B}+4\cdot y_{B}^{2}$

If we know that $x_{A} = 5$ , $y_{A} = 6$ , $x_{B} = 1$ , $y_{B} = b$ , $x_{C} = 1$ and $y_{C} = -3$ , then we have the following expression:

$1 -10 +25 +b^{2} -12\cdot b+36 = 100 -8 +4 +36+24\cdot b +4\cdot b^{2}$

$b^{2}-12\cdot b +52 = 4\cdot b^{2}+24\cdot b +132$

$3\cdot b^{2}+36\cdot b +80 = 0$

This is a second order polynomial, which means the existence of two possible real solutions. By Quadratic Formula, we have the following y-coordinates for point B:

$b_{1} \approx -2.944$ , $b_{2} \approx -9.055$