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frosja888 [35]
2 years ago
6

(a+b)^3 what if it is was a plus ​

Mathematics
2 answers:
bixtya [17]2 years ago
8 0

(a + b)^{3}  =  {a}^{3}  +  3{a}^{2} b + 3a {b}^{2}  +  {b}^{3}

Also do not get mixed between

{a}^{3}  +  {b}^{3}

They both are not the same.

(a + b)^{3} ≠ {a}^{3}  +  {b}^{3}

ankoles [38]2 years ago
7 0

Answer:

Step-by-step explanation:

(a+b)³

a³+a²b+ab²+b³

Or

a³+b³+ab(a+b)

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Pentagon JKLMN is similar to pentagon UVWXY.
marusya05 [52]

9514 1404 393

Answer:

  C.  12

Step-by-step explanation:

Corresponding sides are proportional. Sides with the same mark are the same length.

  WX/XY = LM/MN = LM/KJ

  WX/10 = 18/15

  WX = 180/15 . . . . multiply by 10

  WX = 12

6 0
2 years ago
24h-36? Need helppp please
FinnZ [79.3K]
The trick question is that you can’t effect an unknown value because you don’t know what number it is, it has to be a known value, there is nothing that you can do to figure it out, so it’s unknown and cannot be effected.
7 0
2 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
Pls help me find the area of compound shapes?!?!
Alex73 [517]
3x7 + 2x3 = 22 + 6 = 28 is the area
6 0
2 years ago
Read 2 more answers
Find the area of the figure.
sleet_krkn [62]
Answer=
426cm squared
3 0
3 years ago
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