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3 years ago

FIND AREA OF SHADED SEGMENT. WILL GIVE BRAINLIEST FOR CORRECT ANSWER

Mathematics

2 answers:

Xelga [282]3 years ago

3 0

By geometry,
$area \: of \: shaded \: region \: = (\: area \: of \: square \: - \: 2 \times (area \: of \: semi - circles)) \div 2$
$= ({l}^{2} - 2 \times \pi \times {r}^{2} ) \div 2$
$l = 6 \\ r = l \div 2 = 3$
$= (36 - \pi \times {3}^{2} ) \div 2$
$= 3.86$
area of shaded region = 3.9

Katen [24]3 years ago

3 0

Answer is 3.87
1.First find entire square which is 36
2.Next find semi circle which is 14.13
3.Now multiply by two which is 28.26
4.Now subtract from 36 which is 7.74
5.Divide by 2 and your final answer is 3.87 sq inches

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An equation is an expression used to show the relationship between two or more variables and numbers.

Let x represent the cucumber and y represent the amount of carrots, hence:

20x + 15y = 2(60)

20x + 15y ≤ 120 (1)

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5x + 2y ≤ 25 (2)

The system of 5x + 2y ≤ 25 is given by 20x + 15y ≤ 120 and 5x + 2y ≤ 25

find out more on Equation at: brainly.com/question/2972832

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2 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

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

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

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3 0

3 years ago

Over which interval is the graph of f(x) = –x2 + 3x + 8 increasing?

vivado [14]

Hello there!

To find the increasing intervals for this graph just based on the equation, we should find the turning points first.

Take the derivative of f(x)...
f(x)=-x²+3x+8
f'(x)=-2x+3

Set f'(x) equal to 0...
0=-2x+3
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3/2=x

This means that the x-value of our turning point is 3/2. Now we need to analyze the equation to figure out the end behavior of this graph as x approaches infinity and negative infinity.
Since the leading coefficient is -1, as x approaches ∞, f(x) approaches -∞ Because the exponent of the leading term is even, the end behavior of f(x) as x approaches -∞ is also -∞.

This means that the interval by which this parabola is increasing is...
(-∞,3/2)

PLEASE DON'T include 3/2 on the increasing interval because it's a turning point. The slope of the tangent line to the turning point is 0 so the graph isn't increasing OR decreasing at this point.

I really hope this helps!
Best wishes :)

7 0

3 years ago

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