The length of a rectangle is six times its width. If the rectangle is 600 yd^2 find its perimeter

PLEASE SOMEONE HELP ME OUT which would be the correct equations and how do i solve it? I'll Mark the brainlelist !

sp2606 [1]

Answer:

4 1/4

4 2/8

Step-by-step explanation:

1 7/8 + 2 3/8 = 4 1/4

4 1/4 = 4 2/8

7 0

4 years ago

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SOLVE FOR BRAINLIEST PLEASE

grandymaker [24]

Answer:

x = 20

Explanation:

all the interior angles in the triangle is 180°

therefore:

4x - 17° + 71° + 46° = 180°

4x + 100° = 180°

4x = 180° - 100°

4x = 80°

x = 20°

3 0

2 years ago

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Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

Select two ratios that are equivalent to what are ratios are equivalent to 2:9

finlep [7]

Answer:

3,18 4,27

Step-by-step explanation:

6 0

3 years ago

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In a math class, 12 people reported their grades on the last test. Here are their responses:

Murrr4er [49]

Answer:

(a) 23, 35, 28, 33, 5, 12, 40, 25, 20, 18, 1, 16

1, 5, 12, 16, 18, 20 23, 25, 28, 33, 35, 40

n = 12 M = (20 + 23)/ 2 = 21.5

Q1 = (12 + 16) / 2 = 14 Q3 = (28 + 33) /2 = 30.5

(b) 22, 33, 25, 28, 5, 12, 35, 23, 20, 18, 1, 40, 16

1, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 33, 35, 40

n = 13 M = 22 Q1 = 14 Q3 = 30.5

(c) 20, 28, 23, 25, 3, 5, 33, 22, 18, 16, 40, 1, 35, 12

1, 3, 5, 12, 16, 18, 20, 22, 23, 25, 28, 33, 35, 40

n = 14 M = (20 + 22)/2 = 21 Q1 = 12 Q3 = 28

(d) 20, 28, 23, 25, 3, 5, 30, 22, 18, 40, 16, 35, 1, 33, 12

1, 3, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 30, 33, 35, 40

n = 15 M = 22 Q1 = 12 Q3 = 30

hope this helps :)

5 0

3 years ago