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3 years ago

15

Find sin(90°). 45° A.0 В.1 C. square 2 D. square 2 / 2

1 answer:

otez555 [7]3 years ago

6 0

Answer:

sin90° = 1.

Step-by-step explanation:

Using the diagram, we notice that the side opposite of the 90° angle is the hypotenuse itself.

Therefore sin90° = Hypotenuse/Hypotenuse = 1.

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Refer To The Equation 2x-6y=12<br> Please Help :)

Juli2301 [7.4K]

I'm sorry I can't answer that but I know something so you can answer that. search in browser 'quickmath' all your problems, equations and more can be solved

Step-by-step explanation:

I hope it helps trust me

8 0

3 years ago

If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

Nataly_w [17]

Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

-----------------------------------------------------------------------

Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

4 0

3 years ago

Using the segment addition postulate, which is true?

gulaghasi [49]

The answer is BC + CD = BD

7 0

3 years ago

Read 2 more answers

What is the solution to the system of equations that is graphed?

konstantin123 [22]

Answer:

(-4,-3)

Step-by-step explanation:

Find the point where the 2 lines intersect. In this problem it is (-4,-3)

x=-4

y=-3

8 0

3 years ago

Please help! will mark brainliest

marin [14]

The correct answer is D.

x= -4 and y=2

left 4 up 2

7 0

3 years ago