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stira [4]
3 years ago
9

2z + 4x + 9y -y + 3

Mathematics
1 answer:
kotegsom [21]3 years ago
8 0

Answer:

Step-by-step explanation:

2z+4x+9y-y+3

2z+4x+8y+3

=2z+4x+8y+3

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Any help, please! <br> I dont understand
frutty [35]
The answer would be 40% :)
3 0
3 years ago
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
What is an equation of the line that passes through the point (-6, 1) and has a slope of 3?
AlladinOne [14]
Y= mx +b ; x is the slope and y is the y-intercept.

So in this case, our equation will be y= 3x + 19 :) 
8 0
3 years ago
Writing all Solving Linear Equations in Ule vaDe
zhannawk [14.2K]
P=6.50*10-18.00
hope this helped!❤️
3 0
3 years ago
Read 2 more answers
Express the equation y=2/3x-6 in a standard format
grin007 [14]
Y = 2/3x - 6
-2/3x + y = -6...multiply everything by -3
2x - 3y = 18 <== standard form
3 0
3 years ago
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