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Stella [2.4K]
2 years ago
12

40 POINTS!!!!

Mathematics
1 answer:
Ierofanga [76]2 years ago
5 0

Answer:

It's 1/10

Step-by-step explanation:

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What is |5+3w|/-2 -1
madreJ [45]

Answer:

Step-by-step explanation:

Your maths problem

|

5

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3

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−

2

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1

5 0
2 years ago
Given the function f(x) = -5x + 11 + 3, for what values of x is f(x) = -12?<br> Bleaseeeeee
Nitella [24]

Answer:

x = 5.2

Step-by-step explanation:

f(x) = -5x + 11 + 3

Let's first simplify our first function:

f(x) = -5x + 11 + 3\\f(x) = -5x + 14

Now, we just need to set this function equal to our other given function.

f(x) = -5x+14\\f(x) = -12\\-5x + 14 = -12

Now, we can add 12 to both sides:

-5x + 14 + 12 = -12 + 12\\-5x + 26 = 0

Now, add 5x to both sides:

-5x+26 = 0\\26 = 5x\\

Finally, we'll divide both sides by 5:

26/5 = 5x / 5\\5.2 = x\\x = 5.2

Answer: x = 5.2

6 0
3 years ago
Read 2 more answers
To find the extreme values of a function​ f(x,y) on a curve xequals​x(t), yequals​y(t), treat f as a function of the single vari
pychu [463]

Answer:

Absolute maximum is 2  

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

x=2\cos t,y=2\sin t

By the chain rule:

f'(t)=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }

\frac{df}{dt} =\frac{2 \cos t}{-2\sin t} =-\cot(t)

At fixed points, f'(t)=0

\implies -\cot (t)=0

This gives t=\frac{\pi}{2} ,\frac{3\pi}{2} on 0\le t\le 2\pi

This implies that the extreme points are (2\cos \frac{\pi}{2}, 2\sin \frac{\pi}{2})=(0,2) and (2\cos \frac{3\pi}{2}, 2\sin \frac{3\pi}{2})=(0,-2)

By eliminating the parameter, we have x^2+y^2=4

This is a circle with radius 2, centered at the origin.

Hence (0,2) is an absolute maximum ,at t=\frac{\pi}{2} and (0,-2) is an absolute minimum at  t=\frac{3\pi}{2}

7 0
3 years ago
Find two unit vectors orthogonal to both given vectors. i j k, 4i k
Maurinko [17]
The cross product of two vectors gives a third vector \mathbf v that is orthogonal to the first two.

\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k

Normalize this vector by dividing it by its norm:

\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)

To get another vector orthogonal to the first two, you can just change the sign and use -\mathbf v.
6 0
3 years ago
What is 3a squared add 7 a
vfiekz [6]
The answer is:  " 9a² + 7a " .
_________________________________
Note:    (3a)² + 7a = (3a)*(3a) + 7a = 9a² + 7a .
_________________________________
3 0
3 years ago
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