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makvit [3.9K]
3 years ago
9

[NO LINKS]

Mathematics
2 answers:
Lesechka [4]3 years ago
8 0

Answer:

  • No

Step-by-step explanation:

  • x² – 8x – 9 = 0
  • x² – 8x = 9
  • x² – 8x + 16 = 9 + 16
  • (x – 4)² = 25
  • x – 4 = 25 and x – 4 = - 25 ⇒ Incorrect, should be x - 4 = 5 and x - 4 = -5
  • x = 29 and x = -21                ⇒ Incorrect, should be x = 9 and x = -1
earnstyle [38]3 years ago
7 0

Answer:

No, it's wrong

{x}^{2}  - 8x - 9 = 0 \\ (x - 9)(x + 1) = 0 \\ x = 9 \: and \:  - 1

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The answer is X=5,-13
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Find 2 common angles that sum to (17pi/12) 2. Evaluate tan(17pi/12) using the sum identity for tangent.
Dmitry [639]
Note that
\frac{17 \pi }{12} = \frac{3 \pi }{12} + \frac{14 \pi }{12} = \frac{ \pi }{4} + \frac{7 \pi }{6}

Note that
x= \frac{ \pi }{4}:\,\, sin(x) =cos(x)= \frac{1}{ \sqrt{2} },\,tan(x)=1\\x= \frac{7 \pi }{6} :\,\,sin(x)=- \frac{1}{2} ,\,\,cos(x)=- \frac{ \sqrt{3} }{2} ,\,\,tan(x)= \frac{1}{ \sqrt{3} }

Use the identity
tan(x+y)= \frac{tan(x)+tan(y)}{1-tan(x)tan(y)}

Therefore
tan( \frac{17 \pi }{12} )= \frac{1+ \frac{1}{ \sqrt{3} } }{1- \frac{1}{ \sqrt{3} } } = \frac{ \sqrt{3}+1 }{ \sqrt{3}-1} =  \frac{( \sqrt{3}+1 )^{2}}{( \sqrt{3}-1 )( \sqrt{3}+1 )}  = \frac{3+1+2 \sqrt{3}}{3-1} =2+ \sqrt{3}

Answer: 2 + √3
8 0
3 years ago
Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

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P=-30h+74 The variable h represents the number of hours he has spent reading, the variable p represents the number of pages left
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Answer: Ethan will have to spend 2 hours reading if 14 pages are left to read

Step-by-step explanation:

Given the equation

P=-30h+74

where p =number of pages left to read and

h= the number of hours he has spent reading

Therefore , if Ethan has 14 pages left to read, then the number of hours Ethan will have to read will be

P=-30h+74

14= -30h + 74

30h= 74-14

30h= 60

h = 60/30

h =2 hours

Ethan will have to spend 2 hours reading if 14 pages are left to read

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It’s b because if Mandy goes 18 days and her friend goes 3 times the days she goes.
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