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Ymorist [56]
3 years ago
10

Please help with this ​

Mathematics
1 answer:
levacccp [35]3 years ago
3 0

Step-by-step explanation:

Thats not maths btw

.....

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Are the graphs of each pair of equations parallel, perpendicular, or neither? x=4 y=4
Vlad1618 [11]

Both graphs x =4 and y =4 will be perpendicular to each other.

  • The graph of x = 4 is a vertical line down the Graph scope
  • The graph of y =4 is a horizontal line side the Graph scope.

Both the Lines will meet exactly at ( 4, 4 ) perpendicularly at each other both of them intersecting at 90 degrees.

Vertical and Horizontal lines are perpendicular to each other (90 Degrees).

Know more about Perpendicular Lines: brainly.com/question/1202004

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HELP PLEASE !!!! Rosana used linear combination to solve the system of equations shown. She did so by multiplying the second equ
Allisa [31]
She must have multiplied it by 5.

That would've given her:

x - y = 15
2.5x + y = 25

Now we can add them and the y-terms will eliminate each other. Because one is -y, and the other is positive y. -y + y = 0.
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Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (
natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

I=\int\limits^{}_{} \int\limits^{}_D {x*y}  \, dA =  \int\limits^{}_{} \int\limits^{}_D {x*y}  \, dx*dy

since D is the rectangle such that 0<x<3 , 0<y<3

I=\int\limits^{}_{} \int\limits^{}_D {x*y}  \, dA =  \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y}  \, dx*dy =  \int\limits^{3}_{0} {x}  \, dx\int\limits^{3}_{0} {y}  \, dy  = x^{2} /2*y^{2} /2 =  (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4

4 0
3 years ago
Nina can ride her bike 63,360 feet in 3,400 seconds, and Sophia can ride her bike 10 miles in 1 hour. What is Nina's rate in mil
Sloan [31]
Nina's rate is 12.7 mph and Sophia's rate is 10 mph. Therefore, Nina is faster.
5 0
3 years ago
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PLEASE HELP 25 POINTS!!
eduard

Answer:

(27 y^(-8))/(4 x^6)

Step-by-step explanation:

Simplify the following:

(4 (3 x^2 y^4)^3)/(2 x^3 y^5)^4

Multiply each exponent in 2 x^3 y^5 by 4:

(4 (3 x^2 y^4)^3)/(2^4 x^(4×3) y^(4×5))

4×5 = 20:

(4 (3 x^2 y^4)^3)/(2^4 x^(4×3) y^20)

4×3 = 12:

(4 (3 x^2 y^4)^3)/(2^4 x^12 y^20)

2^4 = (2^2)^2:

(4 (3 x^2 y^4)^3)/((2^2)^2 x^12 y^20)

2^2 = 4:

(4 (3 x^2 y^4)^3)/(4^2 x^12 y^20)

4^2 = 16:

(4 (3 x^2 y^4)^3)/(16 x^12 y^20)

Multiply each exponent in 3 x^2 y^4 by 3:

(4×3^3 x^(3×2) y^(3×4))/(16 x^12 y^20)

3×4 = 12:

(4×3^3 x^(3×2) y^12)/(16 x^12 y^20)

3×2 = 6:

(4×3^3 x^6 y^12)/(16 x^12 y^20)

3^3 = 3×3^2:

(4×3×3^2 x^6 y^12)/(16 x^12 y^20)

3^2 = 9:

(4×3×9 x^6 y^12)/(16 x^12 y^20)

3×9 = 27:

(4×27 x^6 y^12)/(16 x^12 y^20)

4/16 = 4/(4×4) = 1/4:

(27 x^6 y^12)/(4 x^12 y^20)

Combine powers. (27 x^6 y^12)/(4 x^12 y^20) = (27 x^(6 - 12) y^(12 - 20))/4:

(27 x^(6 - 12) y^(12 - 20))/4

6 - 12 = -6:

(27 x^(-6) y^(12 - 20))/4

12 - 20 = -8:

Answer: (27 y^(-8))/(4 x^6)

5 0
4 years ago
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