All categories

3 years ago

13.24503311258278 Round to the nearest tenth as needed

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

7 0

The number provided rounded to the nearest tenth is 13.2.

Because the number in the hundreds place is a 4, we round down and keep the tenth at 2.

You might be interested in

Which of the following could be used to calculate the area of a sector in the circle shown below

lara [203]

Answer:

Therefore the Last option is correct

$\textrm{Area of Sector ACD}=\pi (8\ in)^{2}\times (\frac{42}{360})$

Step-by-step explanation:

Given:

Radius = r = 8 in

θ = 42°

To Find:

Area of Sector = ?

Solution:

We know that

$\textrm{Area of Sector}=\pi (Radius)^{2}\times \frac{\theta}{360}$

Substituting the given values in the formula we get

$\textrm{Area of Sector ACD}=\pi (8\ in)^{2}\times (\frac{42}{360})$

Which is the required Answer.

Therefore the Last option is correct

$\textrm{Area of Sector ACD}=\pi (8\ in)^{2}\times (\frac{42}{360})$

8 0

2 years ago

Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. In y

bekas [8.4K]

F(x) = 2^x; h(x) = x^3 + x + 8

Table

x f(x) = 2^x    h(x) = x^3 + x + 8

0 2^0 = 1 0 + 0 + 8 = 8

1 2^1 = 2    1^3 + 1 + 8 = 10

2 2^2 = 4     2^3 + 2 + 8 = 8 + 2 + 8 = 18

3 2^3 = 8       3^3 + 3 + 8 = 27 + 3 + 8 = 38

4    2^2 = 16    4^3 + 4 + 8 = 76

10    2^10 = 1024 10^3 +10 + 8 = 1018

9    2^9 =   512 9^3 + 9 + 8 = 729 + 9 + 8 = 746

Answer: an approximate value of 10

7 0

3 years ago

Find the 2nd Derivative:<br> f(x) = 3x⁴ + 2x² - 8x + 4

ad-work [718]

Answer:

$f''(x)=36x^2+4$

Step-by-step explanation:

Let's start by finding the first derivative of $f(x)= 3x^4+2x^2-8x+4$ . We can do so by using the power rule for derivatives.

The power rule states that:

$\frac{d}{dx} (x^n) = n \times x^n^-^1$

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

$\frac{d}{dx} (3x^4+2x^2-8x+4)$

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)$

Simplify the equation.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)$
$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)$

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)$
$f'(x)=12x^3+4x-8$

Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).