49=16+x^2
x^2= 33
x= square root of 33
The correct answer is: [B]: " (2, 5) ".
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Given:
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-5x + y = -5 ;
-4x + 2y = 2 .
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Consider the first equation:
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-5x + y = -5 ; ↔ y + (-5x) = -5 ;
↔ y - 5x = -5 ; Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
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y - 5x + 5x = -5 + 5x
y = -5 + 5x ; ↔ y = 5x - 5 ;
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Now, take our second equation:
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-4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ; and solve for "x" :
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-4x + 2(5x - 5) = 2 ;
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Note, 2(5x - 5) = 2(5x) - 2(5) = 10x - 10 ;
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So: -4x + 10x - 10 = 2 ;
On the left-hand side of the equation, combine the "like terms" ;
-4x +10x = 6x ; and rewrite:
6x - 10 = 2 ;
Now, add "10" to each side of the equation:
6x - 10 + 10 = 2 + 10 ;
to get:
6x = 12 ; Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;
6x/6 = 12 / 6 ;
x = 2 ;
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Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
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-5x + y = -5 ;
-5(2) + y = -5 ;
-10 + y = -5 ; ↔
y - 10 = -5 ;
Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;
y - 10 + 10 = -5 + 10 ;
y = 5 .
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So, we have, x = 2 ; and y = 5 .
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Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
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first equation:
-5x + y = -5 ;
-5(2) + 5 =? -5?
-10 + 5 =? -5 ? YES!
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second equation:
-4x + 2y = 2 ;
-4(2) + 2(5) =? 2 ?
-8 + 10 =? 2 ? Yes!
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So, the answer is:
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x = 2 , y = 5 ; or, "(2, 5)" ; which is: "Answer choice: [B] " .
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<span>For the plane, we have z = 5x + 9y
For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals
![\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx \\ \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx \\ \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\ \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E1_0%20%7B%20%5Cint%5Climits_%7Bx%5E4%7D%5Ex%20%7B%285x%2B9y%29%7D%20%5C%2C%20dy%20%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B5xy%2B%20%5Cfrac%7B9%7D%7B2%7D%20y%5E2%5Cright%5D_%7Bx%5E4%7D%5E%7Bx%7D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B%5Cleft%285x%28x%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%29%5E2%5Cright%29-%5Cleft%285x%28x%5E4%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%5E4%29%5E2%5Cright%29%5Cright%5D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%285x%5E2%2B%20%5Cfrac%7B9%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%20%5Cfrac%7B19%7D%7B6%7D%20x%5E3-%20%5Cfrac%7B5%7D%7B6%7D%20x%5E6-%20%5Cfrac%7B1%7D%7B2%7D%20x%5E9%5Cright%5D%5E1_0)
</span>
The answer would be A because each side is being multiplied by 3
Answer:
Step-by-step explanation:
<u>This is a linear relationship y= mx + b and:</u>
- The y-intercept is b = 12.60
- The slope is m = 0.64
<u>The monthly total is:</u>
