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3 years ago

Please help me out with this :)

Mathematics

2 answers:

aleksandr82 [10.1K]3 years ago

8 0

Answer:

c=0

Step-by-step explanation:

*graphed*

coordinates are 0,0 6,8 0,12 10,0

*plug them into c=x+y*

C=0+0=0

C=6+8=14

C=0+12=12

C=10+0=10

since 0 is the smallest value C=o and if they want a fr number then your second option is C=10

valentinak56 [21]3 years ago

3 0

Answer:

C = 0 is the minimum value

Step-by-step explanation:

Sketch

2x + y = 20

with x- intercept = (10, 0) and y- intercept = (0, 20)

2x + 3y = 36

with x- intercept = (18, 0) and y- intercept = (0, 12)

Solve 2x + y = 20 and 2x + 3y = 36 to find intercept at (6, 8)

The feasible region has vertices at

(0, 20), (12, 0) and (6, 8)

Evaluate the objective function C = x + y at each vertex

(0, 20) → C = 0 + 20 = 20

(12, 0) → C = 12 + 0 = 12 ← minimum value

(6, 8) → C = 6 + 8 = 14

The minimum value is C = 12 when x = 12 and y = 0

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3(3t + s) and 3s - t S = 9 and t =12

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4 years ago

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What is the solution to the inequality -3x - 42 > 3

garik1379 [7]

$- 3x - 42 > 3\Leftrightarrow - 3x > 45\Leftrightarrow x < - 15$

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4 years ago

The following three shapes are based only on squares, semicircles, and quarter circles. Find the perimeter and the area of each

bixtya [17]

When we divide the figure in four parts, we obtain four squares with its sides of 5 centimeters of lenght, with quarter circles (two) in each one of them. The limits of the shadded area are arcs of cirncunference

To calculate the area of the shadded part, we must choose one square of 5cmx5cm and divide it in 3 sectors:

a: the area below the shaded area.
b: the area above the shaded area.
c: the shaded area.

The area of the square (As) is:
As=L²
As=(5 cm)²
As=25 cm²

The area of a circunference is: A=πR² (R:radio), but we want the area of the quarter circle, so we must use A=1/4(πR²), to calculate the area of the sectors a+c:
A(a+c)=1/4(π(5)²)
A(a+c)=25π/4

The area of the sector "b" is:
Ab=As-A(a+c)
Ab=25-25π/4
Ab=25(1-π/4)

The area of the sector a+b, is:
A(a+b)=2Ab
A(a+b)=2x25(1-π/4)
A(a+b)=50(1-π/4)

Then, the shadded area (Sector c) is:
Ac=As-A(a+b)
Ac=25-50(1-π/4)
Ac=25-(50-50π/4)
Ac=25-50+50π/4
Ac=(50π/4)-25
Ac=(25π/2)-25
Ac=25(π/2-1)

The area of each shaded part is: 25(π/2-1)

To calculate the perimeter of a shaded part, we must remember that the perimeter of a circunference is: P=2πR. If we want the perimeter of a quarter circle we must use: P= 2πR/4. But there is two quarter circles in the square of 5cmx5cm, so the perimeter of the shaded area is:

P=2(2πR/4)
P=4πR/4
P=πR
P=5π

The perimeter of each shaded part is: 5π

6 0

3 years ago

In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the d

nikdorinn [45]

Answer:

R = 7,34 km

Ф = 63,43° north of east

Step-by-step explanation:

Let´s analyze each movement and find its perpendicular components. Then, we are going to add the components along the x and y axes to get the resultant and finally to calculate its magnitude and direction.

To determine the perpendicular components, we will use right triangles trigonometrics ratios.

Please see images for each displacement in attached file.

Displacement A: 2.50 km and 45° north of west

Ax= AcosФ = 2.50cos45° = 1.77 km (neg. direction)

Ay= AsenФ = 2.50sen45° = 1.77 km (pos. direction)

Displacement B: 4.70 km and 60° south of east

Bx= BcosФ = 4.70cos60° = 2,35 km (pos. direction)

By= BsenФ = 4.70sen60° = 4,07 km (neg. direction)

Displacement C: 1.30 km and 25° south of west

Cx= CcosФ = 1.30cos25° = 1,18 km (neg. direction)

Cy= CsenФ = 1.30sen25° = 0.55 km (neg. direction)

Displacement D: 5.10 km due east (0°)

Dx= DcosФ = 5.10 km (pos. direction)

Dy= DsenФ = 0 km

Displacement E: 1.70 km and 5° east of north

Ex= EsenФ = 1.70sen5° = 0,15 km (pos. direction)

Ey= EcosФ = 1.70cos5° = 1.69 km (pos. direction)

Displacement F: 7.20 km and 55° south of west

Fx= FcosФ = 7.20cos55° = 4.13 km (neg. direction)

Fy= FsenФ = 7.20sen55° = 5.90 km (neg. direction)

Displacement G: 2.80 km and 10° north of east

Gx= GcosФ = 2.80cos10° = 2.76 km (pos. direction)

Gy= GsenФ = 2.80sen10° = 0.49 km (pos. direction)

Now, we add the components along the x- and y- axis to find the components of the resultant (R):

Rx = Ax + Bx + Cx + Dx + Ex + Fx + Gx

Ry = Ay + By + Cy + Dy + Ey + Fy + Gy

Therefore,

Rx= (-1.77)+(2.35)+(-1.18)+(5.10)+(0.15)+(-4.13)+(2.76)

Ry= (1.77)+(-4.07)+(-0.55)+(0)+(1.69)+(-5.90)+(0.49)

Note: minus (-) symbol to negative directions

Rx = 3.28 km

Ry = - 6.57 km

Let´s use the Theorem of Pythagoras to find the magnitud ot the resultant:

$R = \sqrt{Rx^{2} +Ry^{2} }$

$R =\sqrt{(3.28)^{2}+(-6.57)^{2} }$

R = 7,34 km

To find the direction of the resultant:

tanФ = $\frac{Ry}{Rx}$

tanФ = $\frac{6.57}{3.28}$

tanФ = 2.00

Ф = tan-1 (2.00)

Ф = 63,43° north of east

Download pdf

3 0

4 years ago

What is the center of the circle and the radius of 2x^2 - 12x +2y^2 +8y = 10

DedPeter [7]

Answer:

the center of the circle, (h, k), is (3, -2),

and the radius is √18, or 3√2.

Step-by-step explanation:

This problem becomes much easier if you divide it through by 2:

x^2 - 6x +y^2 +4y = 5

Next, regroup this equation so that the x terms are together and the y terms are also together, separately:

x^2 - 6x + y^2 + 4y = 5

Next, complete the square of x^2 - 6x:

x^2 - 6x → x^2 - 6x + 9 - 9 → (x - 3)^2 - 9.

Then complete the square of y^2 + 4y:

y^2 + 4y + 4 - 4 → (y + 2)^2 - 4

Now rewrite x^2 - 6x + y^2 + 4y = 5 as

(x - 3)^2 - 9 + (y + 2)^2 - 4 = 5

Next, group the constants together on the right side:

(x - 3)^2 - 9 + (y + 2)^2 - 4 = 5 → (x - 3)^2 + (y + 2)^2 = 18

Comparing this result to the standard equation of a circle:

(x - h)^2 + (y - k)^2 = r^2,

we see that h = 3, y = -2 and r^2 = 18.

Thus, the center of the circle, (h, k), is (3, -2),

and the radius is √18, or 3√2.

6 0

3 years ago