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1 year ago

Show that the equation x^3+7x-4=0 has a solution between x=0 and x=1

Mathematics

1 answer:

Aleksandr [31]1 year ago

3 0

Step-by-step explanation:

Using the Intermediate Value Theorem, the following is applied:

"If f(x) is a continuous on interval [a,b] and we have two points f(a) and f(b) then there must be some value c such that f(a)<f(c)<f(b).

So here there must be a c such that

$f(0) < f(c) < f(1)$

Note: F(c)=0, the questions that the function have a solution between 0 and 1, so that means we must have some value, c such that f(c)=0 that exists

Next, plug in the x values into the function

$f(0) = - 4$

$f(1) = {1}^{3} + 7(1) - 4 = 4$

Since cubic functions are continuous and -4<0<4, then there is a solution c that lies between f(0) and f(1)

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which functions have a range of all real numbers ? check all that apply. A. y=csc x B. y=sec x C. y=tan x D. y= cot x

Natalka [10]

None of these. All of these functions are defined as ratio of trigonometric functions.

Trigonometric functions have infinite zeroes, so when you put them in the denominator, they lead to infinitely many points of ill-definition.

Specifically, we have:

$\csc(x) = \dfrac{1}{\sin(x)}$

which is undefined at

$\sin(x)=0\iff x=k\pi, k\in\mathbb{Z}$

$\sec(x) = \dfrac{1}{\cos(x)}$

which is undefined at

$\cos(x)=0\iff x=\dfrac{\pi}{2}+k\pi, k\in\mathbb{Z}$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$

which is undefined at

$\cos(x)=0\iff x=\dfrac{\pi}{2}+k\pi, k\in\mathbb{Z}$

$\cot(x) = \dfrac{\cos(x)}{\sin(x)}$

which is undefined at

$\sin(x)=0\iff x=k\pi, k\in\mathbb{Z}$

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2 years ago

Calculate the perimeter of the figure shown below.

Julli [10]

Answer:

Step-by-step explanation:

perimeter means that you add up all the sides do the perimeter is 41

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2 years ago

Does the table represent a function? Why or why not?

Svetradugi [14.3K]

Answer:

Yes it does

Step-by-step explanation:

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2 years ago

20. what us the area of the figure:

Varvara68 [4.7K]

If the angle in B is 45º, that means the angle in A is also going to be 45º, considering it's a right triangle, so AC and BC are they have the same length.
Then using Pythagoras, you'll get that $AC^{2} + BC^{2}$ equals $AB^{2}$ .
Now, you know that AC=BC and that AB=24.
So you'll get $2AC^{2} =24^2$ . You do the square root in both sides and you get that 2AC=24 and AC=12.

Now that you know that both AC and BC equal 12, you can find the area by just multiplying them and then diving them by 2. (The formula for the area of a triangle is half base multiplied by height, and in a right triangle, if a cathetus is the base, the other cathetus<span> will be the height)

And so the area is equal to 72.</span>

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3 years ago

Use integration by parts to find the integrals in Exercise.<br> ∫(4x-12)e-8x dx.

stepladder [879]

Answer:

$e(2x^{2} -12x)-4x^{2}+C$