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ioda
3 years ago
15

(2+x)(2−x)?? helppppppp

Mathematics
2 answers:
shusha [124]3 years ago
3 0
You can find one the picture I send




il63 [147K]3 years ago
3 0

Answer:

-x^2+4

Step-by-step explanation:

2*2=4

2*-x=-2x

x*2=2x

x*-x=-x^2

-x^2+2x-2x+4=-x^2+4

          \   /

      These two

      cancel out

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WILL MARK BRAINLIST PLS HELP:))
dusya [7]

Answer:

Vertex: (3,0)

Max/min: min

axis of symmetry: x=3

Domain: (-∞,∞)

Range: [3,∞)

zeroes: (3,0)

Step-by-step explanation:

Vertex is where the graph changes directions (so in this case it's the point where it changes from decreasing to increasing) which I think is (3,0)

It's a minimum because the coefficent for the degree is positive

The axis of symmetry is just the x value of the vertex (which is x= 3)

the domain is all possible x values (-∞,∞)

The range is all possible y values [3,∞)

The zeroes is where the line hits the x axis, which is (3,0)

6 0
3 years ago
PLEASE HELP! Brainliest answer for first correct answer...
trasher [3.6K]
        check picture below

8 0
3 years ago
16=5y/3<br><br> HELLLPPP PLSSSS
Lunna [17]

Answer:

y = 9.6

Step-by-step explanation:

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4 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
Given h(x) = 5x + 2, solve for 2 when h(x) = -8.
san4es73 [151]

Answer:

\huge\boxed{\sf x = -2}

Step-by-step explanation:

\sf h(x) = 5x+2\\\\Put \ h(x) = -8\\\\-8 = 5x+2\\\\Subtract \ 2 \ to \ both \ sides\\\\-8-2 = 5x\\\\-10 = 5x\\\\Divide\ both \ sides \ by \ 5\\\\-10 / 5 = x \\\\x = -2 \\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
8 0
2 years ago
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