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3 years ago

Help with this math thingy pls! <3

Mathematics

1 answer:

zvonat [6]3 years ago

8 0

45.6 hope this helps

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Problem Solving World

jonny [76]

Answer:

Step-by-step explanation:

To solve this, add.

28 + 18 = 46

7 0

3 years ago

Read 2 more answers

What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?

nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

$f(x)=x^5-10x^3+9x$ ; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

$f'(x)=5x^4-30x^2+9$

Setting derivative equal to 0, we get

$5x^4-30x^2+9=0$

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376 : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = $x^5- 10x^3+ 9x$ on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

7 0

3 years ago

Which equation choice could represent the graph shown below?

BartSMP [9]

It has to be an equation that adds up to an exponent of 3

4 0

2 years ago

An assembly consists of two mechanical components. Suppose that the probabilities that the first and second components meet spec

SVETLANKA909090 [29]

Answer:

The probability mass function is given by

P(X = 0) = 0.003, P(X = 1) = 0.164, P(X = 2) = 0.833

Step-by-step explanation:

Given to us are two mechanical components.

The probability of the first component meeting the given specifications are 0.98.

The probability that the second component meets the given specifications = 0.85.

In the information given the components are said to be independent.

The probability mass function will be calculated in the following way

i.) P(X=0) which is the probability that neither of the components meet the specifications.

ii) P(X=1) which is the probability when either of the components will meet the specifications and the other will not.

iii) P(X=2) which is the probability that both components meet the specification.

Therefore P(X = 0) = ( 1 - 0.98) × ( 1 - 0.85) = 0.02 × 0.15 = 0.003

and, P(X = 1) = (0.98 × ( 1 - 0.85)) + (( 1 - 0.98) × 0.85)

= (0.98 × 0.15) + (0.02 × 0.85)

= 0.147 + 0.017

= 0.164

and finally, P(X=2) = (0.98 × 0.85)

= 0.833

4 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

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