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2 years ago

PLease help me solve this

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

4 0

Answer: 729

Step-by-step explanation:

$\frac{1}{3^{-2} * 3^{-4} * (-1)^{2} }$

3^-2 = 1/9

3^-4 = 1/81

(-1)^2 = 1

1/(1/9 * 1/81 * 1)

1/ 1/729 = 729

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Which equation represents a line which is perpendicular to the line 5y-4x=-15 5y−4x=−15?

Lady bird [3.3K]

Y = -5/4x + b is the equation that represents a line that is perpendicular to the line 5y-4x=-15

If you want it in standard form, it is 5x+4y=-15

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3 years ago

Give an example of a function F(x) that is an antiderivative of f(x) = 9 cos(9x)+ 3x^2

liubo4ka [24]

We are integrating f(x) = 9cos(9x) + 3x²: $\int\ {9cos(9x)+3x^{2} } \, dx$

a) Apply the sum rule

$\int\9cos(9x)} \, dx +\int\ 3x^{2} } \, dx$

b) Calculate each antiderivative

<u>First integral</u>

$\int\ {9cos(9x)} \, dx$

1. Take out the constant

$9\int\ {cos(9x)} \, dx$

2. Apply u-substitution, where u is 9x

$9\int\ {cos(u)\frac{1}{9} } du$

3. Take out the constant (again)

$9*\frac{1}{9} \int{cos(u)} du$

4. Take the common integral of cos, which is sin

$9*\frac{1}{9}sin(u)}$

5. Substitute the original function back in for u and simplify $9*\frac{1}{9} sin(9x) = sin(9x)$

6. Always remember to add an arbitrary constant, C, at the end

$sin(9x) + C$

<u>Second integral</u>

$\int3x^{2} } \, dx$

1. Take out the constant

$3\int{x^{2} } \, dx$

2. Apply the power rule, $\int{x}^{a} \, dx =\frac{x^{a+1} }{a+1}$ , where <em>a</em> is your exponent

⇒ $3*\frac{x^{2+1} }{2+1} = x^{3}$

3. Add the arbitrary constant

$x^{3} + C$

c) Add the integrals

sin(9x) + C + x³ + C = sin(9x) + x³ + C

Notice the two arbitrary constants. Since we do not know what either constant is, we can combine them into one arbitrary constant.

<h3>Answer:</h3>

F(x) = sin(9x) + x³ + C

3 0

2 years ago

Read 2 more answers

The coordinate plane shows the floor plan for a swimming pool. What is the area of the pool's border?

LekaFEV [45]

Answer:

border area: 100 m²

Step-by-step explanation:

The area of the pool border will be the overall area of the pool and boder, less the area of the pool. The area of each rectangle is the product of its length and width. The length and width of each can be found using the Pythagorean theorem.

<h3>Pythagorean Triple</h3>

The triple (3, 4, 5) is called a Pythagorean triple, because ...

3² +4² = 5²

That is, a triangle with sides of lengths 3, 4, and 5 will be a right triangle with a hypotenuse of length 5.

<h3>Application to Pool/Border Dimensions</h3>

We notice that the sides of the rectangles in the figure have a rise/run of ±3/4 or ±4/3. In each case, the side of the rectangle is the hypotenuse of a triangle with dimensions a multiple of 3 or 4 meters. Hence the hypotenuse (rectangle side length) is the corresponding multiple of 5 meters. The attachment illustrates this point, showing a triangle with sides that are (6, 8, 10), 2 times the basic triple.

The pool is 5 m by 10 m, for an area of ...

A = LW = (10 m)(5 m) = 50 m²

The overall area is 10 m by 15 m, for an area of ...

A = (15 m)(10 m) = 150 m²

<h3>Border Area</h3>

The pool's border is the difference between the overall area and the pool area:

150 m² -50 m² = 100 m² . . . . area of pool's border