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3 years ago

I need help in 2 3 4 5 6 7 8

Mathematics

1 answer:

ahrayia [7]3 years ago

7 0

Yo i can't see wut it says

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Whats a multiplication problem with a product of 5 to the 13th power

fredd [130]

5 to the 7th power times 5 to the 6th power

8 0

3 years ago

Explain why the "+ C" is not needed in the antiderivative when evaluating a definite integral.

Aliun [14]

The reason the "+ C" is not needed in the antiderivative when evaluating a definite integral is; The C's cancel each other out as desired.

<h3>How to represent Integrals?</h3>

Let us say we want to estimate the definite integral;

I = $\int\limits^a_b {f'(x)} \, dx$

Now, for any C, f(x) + C is an antiderivative of f′(x).

From fundamental theorem of Calculus, we can say that;

$I = \int\limits^a_b {f'(x)} \, dx = \phi(a) - \phi(b)$

where Ф(x) is any antiderivative of f'(x). Thus, Ф(x) = f(x) + C would not work because the C's will cancel each other.

Read more about Integrals at; brainly.com/question/22008756

#SPJ1

8 0

2 years ago

An open box is to be made from a six inch by six inch piece of material by cutting equal squares from the corners and turning up

Virty [35]

Answer:

8 inch³

Step-by-step explanation:

Side should be 2 inch

Volume = side³ = 2³ = 8 inch³

4 0

3 years ago

Read 2 more answers

Can someone help me pls? Thankss:)

zloy xaker [14]

Answer: C

Step-by-step explanation:

8 0

3 years ago

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$