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What is the value of the expression when a = 6, b = 4, and c = 8? 2a/3b−cA: 1

B: 3

C: 8

D: 12

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

$(2a)/(3b-c) \n=(2(6))/(3(4) - 8) \n=(12)/(12-8) \n=(12)/(4) \n=3$

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What is the next number in the sequence? 3….9….27….81….A) 162
B) 180
C)243
D) 270

Answers

The next number should be 81*3, so 243.

The Point (0,0) is a solution to which of these inequalities?

Answers

Answer:

(0, 0) s the solution of y-4 < 3x-1 as it satisfies the inequality.

Hence, option C is true.

Step-by-step explanation:

Given the point (0, 0)

Putting the point (0, 0) the inequality

y+4 < 3x-1

0+4 < 3(0)-1

4 < -1

This is false as -1 can not be greater than 4

y-1 < 3x-4

Putting the point (0, 0) the inequality

0-1 < 3(0)-4

-1 < -4

This is false as -1 can not be lesser than -4

y-4 < 3x-1

Putting the point (0, 0) the inequality

0-4 < 3(0)-1

-4 < -1

This is true as -4 is lesser than -1

y+4 < 3x+1

Putting the point (0, 0) the inequality

0+4 < 3(0)+1

4 < 1

This is false as 4 can not be lesser than 1

Therefore, (0, 0) s the solution of y-4 < 3x-1 as it satisfies the inequality.

Hence, option C is true.

I do not get question 5.

Answers

Answer:

Take the number shown and divided by 5000

Step-by-step explanation:

Answer:

a) is correct

b) 0.3768

c) 0.5246

Step-by-step explanation:

456 + 1428 = 1884

1884 ÷ 5000 = 0.3768

739 + 456 + 1428 = 2623

2623 ÷ 5000 = 0.5246

(1 point) For the given position vectors r(t)r(t), compute the (tangent) velocity vector r′(t)r′(t) for the given value of tt . A) Let r(t)=(cos4t,sin4t)Let r(t)=(cos⁡4t,sin⁡4t). Then r′(π4)r′(π4)= ( , )? B) Let r(t)=(t2,t3)Let r(t)=(t2,t3). Then r′(5)r′(5)= ( , )? C) Let r(t)=e4ti+e−5tj+tkLet r(t)=e4ti+e−5tj+tk. Then r′(−5)r′(−5)= i+i+ j+j+ kk ?

Answers

Answer:

(a)

$r'(\frac \pi 4) =(0.-4)$

(b)

r'(5)= (10,75)

(c)

$r'(-5) =4 e^(-20)\hat i-5e^(25)\hat j+\hat k$

Step-by-step explanation:

(a)

Give that,the position vector is

r(t) = (cos 4t, sin 4t)

Differentiating with respect to t

r'(t) = (-4sin 4t, 4 cos 4t) [ $(d)/(dt) cos mt = -m \ sin \ mt$ and $(d)/(dt) sin mt = m \ cos \ mt$ ]

To find the $r'(\frac\pi 4)$ , we put $t=\frac \pi4$

$r'(\frac\pi 4) = (-4sin (4.\frac \pi 4), 4 cos (4.\frac \pi 4))$

=(0, -4)

(b)

Give that,the position vector is

r(t) = (t²,t³)

Differentiating with respect to t

r'(t) = (2t, 3t²)

To find r'(5) , we put t=5

r'(5) = (2.5,3.5²)

= (10,75)

(c)

Given position vector is

$r(t) = e^(4t)\hat i+e^(-5t)\hat j+t\hat k$

Differentiating with respect to t

$r'(t) =4 e^(4t)\hat i+(-5)e^(-5t)\hat j+\hat k$

$\Rightarrow r'(t) =4 e^(4t)\hat i-5e^(-5t)\hat j+\hat k$

To find r'(-5) , we put t= - 5 in the above equation

$r'(-5) =4 e^(4.(-5))\hat i-5e^(-5.(-5))\hat j+\hat k$

$\Rightarrow r'(-5) =4 e^(-20)\hat i-5e^(25)\hat j+\hat k$

For the given position vectors r(t)r(t), compute the (tangent) velocity vector r′(t)r′(t) for the given value of tt are:

$A) r' (\pi /4) = (0, -4) \nB) r'(5) = (10, 75)\nC) r'(-5) = (4e^(-20), -5e^(25), 1)$

To compute the velocity vector, we need to find the derivative of the position vector with respect to time (t). This will give us the tangent velocity vector.

A) Let r(t) = (cos⁡4t, sin⁡4t).

To find r'(t), we take the derivative of each component with respect to t:

r'(t) = (d/dt (cos⁡4t), d/dt (sin⁡4t))

r'(t) = (-4sin⁡4t, 4cos⁡4t)

To find r'(π/4), we substitute t = π/4 into r'(t):

r'(π/4) = (-4sin⁡(4(π/4)), 4cos⁡(4(π/4)))

r'(π/4) = (-4sin⁡π, 4cos⁡π)

r'(π/4) = (0, -4)

B) $Let \ r(t) = (t^2, t^3).$

To find r'(t), we take the derivative of each component with respect to t:

$r'(t) = (d/dt (t^2), d/dt (t^3))\nr'(t) = (2t, 3t^2)$

To find r'(5), we substitute t = 5 into r'(t):

$r'(5) = (2(5), 3(5)^2)\nr'(5) = (10, 75)$

C) Let $r(t) = e^(4t)i + e^(-5t)j + tk.$

To find r'(t), we take the derivative of each component with respect to t:

$r'(t) = (d/dt (e^(4t)), d/dt (e^(-5t)), d/dt (t))]\n\nr'(t) = (4e^(4t)), -5e^(-5t), 1)$

To find r'(-5), we substitute t = -5 into r'(t):

$r'(-5) = (4e^(4{-5}), -5e^(-5(-5)), 1) \n\nr'(-5) = (4e^(-20), -5e^(25), 1)$

So, the answers are:

$A) r' (\pi /4) = (0, -4) \nB) r'(5) = (10, 75)\nC) r'(-5) = (4e^(-20), -5e^(25), 1)$

To know more about vectors:

brainly.com/question/33923402

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$(x)/(2) + 10 = * - 3$ The answer is 26, but I dont know how to solve it

Answers

First of all move all the Xs to the left side, and the numbers to the left.
So
x/2-x = -13
Which means
X/2-x/1 = -13
Which means x/2-2x/2 = -13
So u combine to two fractions
-x/2 = -13
-x = -13 * 2
Which means x = 26 (u flip the negativity)

Is 1 greater than 1?