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2 years ago

What are 3 ways you could use arts and music outside school

Mathematics

2 answers:

dimaraw [331]2 years ago

6 0

Painting a mural
Doing charity work like playing a concert
Design

Helen [10]2 years ago

4 0

You can preform in a concert and paint

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8a2 = 12 + a plzzzzzzzzzzzzzzzzzzzzzzzzzz help me slove this problem

polet [3.4K]

Answer:

1.28

Step-by-step explanation:

Assuming in 8a2 the '2' represents an exponent, you answer should be

1.28

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1 year ago

A carpenter has a board 400 inches long and 10 inches wide. He makes 5 identical shelves with the board and still has a piece of

Sedaia [141]

THE 5 IDENTICAL SHELVES WOULD have to be 76 because if you take 400 - 20 (because that left over) you get 380/5 you get 76 and to check your work you do 5x76=380+20 you get 400 so 76 is you answer

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2 years ago

Read 2 more answers

What is the second step in solving this equation by completing the square 3x^2+14x+8=0

Trava [24]

The second step is to keep your x terms on the left and move the constant to the right by subtracting it from both sides.

8 0

2 years ago

EThe graph of a quadratic function passes through the points (6,0) and (p,0). The maximum point

MA_775_DIABLO [31]

Answer:

p = - 5

Step-by-step explanation:

The vertex of the function is (0.5, 30.25 )

Since the roots are (6, 0) and (p, 0) then the axis of symmetry is vertical

with equation x = 0.5

The axis of symmetry passes through the midpoint of the roots, thus

$\frac{6+p}{2}$ = 0.5 ( multiply both sides by 2 )

6 + p = 1 ( subtract 6 from both sides )

p = - 5

6 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

2 years ago