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1 year ago

One approximate solution to the equation cos x = -0.60 for the domain 0° ≤ x ≤ 360° is

Mathematics

1 answer:

hjlf1 year ago

5 0

307 klkammamamamammama

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A 3 x a 4 = a 12

PolarNik [594]

Answer:

3a × 4a = 12 a^2

Step-by-step explanation:

each and every term is multiplied, never leave any term without multiplying it

4 0

3 years ago

Read 2 more answers

Find the maximum value of y = -2(x - 4)2 - 6. A) -6 B) -4 C) -2 D) 4

otez555 [7]

Y=-2(x-4)^2-6

The -6 makes it shift from origin(0,0) to (0,-6)

Then the -4 inside with x, makes it shift to the right by for, so the center would be now instead of (0,0), (4,-6).

As it has a negative A factor, it is a parabola open downwards, so the center is the maximum value.

Answer: (4,-6)

Hope you get it!

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4 years ago

4/5 y = 8 , how do we solve pls help fast its urgent, I will mark u brainliest

Butoxors [25]

Answer:

y = 10

Step-by-step explanation:

4/5 y = 8

4/5 y * 5/4 = 8 * 5/4 multiply by reciprocal

y = 10

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3 years ago

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1. What is the area of the shape?<br> 6cm<br> 4cm<br> 2cm<br> 3cm

Solnce55 [7]

Hello,
I’m not sure how to answer your question because looks like you forgot to put an image of the shape your talking about.

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3 years ago

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

$y'=1-2x$

It can be written as

$\frac{dy}{dx}=1-2x$

Use variable separable method to solve this differential equation.

$dy=(1-2x)dx$

Integrate both the sides.

$\int dy=\int (1-2x)dx$

$y=x-2(\frac{x^2}{2})+C$ $[\because \int x^n=\frac{x^{n+1}}{n+1}]$

$y=x-x^2+C$ ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

$-2=1-(1)^2+C$

$-2=1-1+C$

$-2=C$

The value of C is -2. Substitute C=-2 in equation (1).

$y=x-x^2-2$

Therefore the solution of the given initial value problems in explicit form is $y=x-x^2-2$ .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.

4 0

3 years ago