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Andrew [12]
3 years ago
12

Find the solution of 3 times the square root of the quantity of x plus 6 equals negative 12, and determine if it is an extraneou

s solution.
Mathematics
1 answer:
Anna35 [415]3 years ago
6 0
For this case we have the following equation:
 3 \sqrt{x+6}=-12
 Rewriting we have:
 \sqrt{x+6}= \frac{-12}{3}
 \sqrt{x+6}=-4
 We raise both members of the equation to the square:
 (\sqrt{x+6})^2=(-4)^2
 Rewriting we have:
 x+6=16
 x=16-6
 x=10
 It's a extraneous solution because equality is not met by substituting x = 10 in the original equation:
 3 \sqrt{10+6}=-12
 3 \sqrt{16}=-12
 3(4)=-12
 12=-12
 Answer:
 
The solution is:
 
x=10
 It's a extraneous solution

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Circumference of a circle with radius 10 in
azamat
2*pi*10=62.83 is certainly your answer
5 0
3 years ago
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What is c in 6c + 3y = 18 if y is 2?
Natali [406]

Answer:

c = 2

Step-by-step explanation:

6c + 3y = 18

6c + 3(2) = 18

6c + 6 = 18

6c = 12

c = 2

6 0
2 years ago
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What is the longest line segment that can be drawn in a right rectangular prism that is 15 cm​ long, 11 cm​ wide, and 10 cm​ tal
sammy [17]

Answer:

  √446 ≈ 21.12 cm

Step-by-step explanation:

The longest dimension of a rectangular prism is the length of the space diagonal from one corner to the opposite corner through the center of the prism. The Pythagorean theorm tells you the square of its length is the sum of the squares of the dimensions of the prism:

  d² = (15 cm)² +(11 cm)² +(10 cm)² = (225 +121 +100) cm² = 446 cm²

  d = √446 cm ≈ 21.12 cm

The longest line segment that can be drawn in a right rectangular prism is about 21.12 cm.

_____

<em>Additional comment</em>

The square of the face diagonal is the sum of the squares of the dimensions of that face. The square of the space diagonal will be the sum of that square and the square of the remaining prism dimenaion, hence the sum of squares of all three prism dimensions.

8 0
2 years ago
Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu
Alisiya [41]
Let y=C_1x+C_2x^3=C_1y_1+C_2y_2. Then y_1 and y_2 are two fundamental, linearly independent solution that satisfy

f(x,y_1,{y_1}',{y_1}'')=0
f(x,y_2,{y_2}',{y_2}'')=0

Note that {y_1}'=1, so that x{y_1}'-y_1=0. Adding y'' doesn't change this, since {y_1}''=0.

So if we suppose

f(x,y,y',y'')=y''+xy'-y=0

then substituting y=y_2 would give

6x+x(3x^2)-x^3=6x+2x^3\neq0

To make sure everything cancels out, multiply the second degree term by -\dfrac{x^2}3, so that

f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y

Then if y=y_1+y_2, we get

-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0

as desired. So one possible ODE would be

-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0

(See "Euler-Cauchy equation" for more info)
6 0
3 years ago
A ________ relationship indicates that for each record in a table, there is only a single corresponding record in a related tabl
Maurinko [17]
I think its a one-to-one relationship.
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3 years ago
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