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svp [43]
3 years ago
13

The expression on the left side of an equation is show below

Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

−4(x−2)+5x

Distribute:

=(−4)(x)+(−4)(−2)+5x

=−4x+8+5x

Combine Like Terms:

=−4x+8+5x

=(−4x+5x)+(8)

=x+8

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Identify the polynomial 4x3 + 6x2
Leona [35]

Answer:

4 x 3 + 6 x 2 has 4 terms so it's an quadrinomial

Step-by-step explanation:

D quadrinomial

5 0
3 years ago
What is the answer to 2/3 (x +2) -1/3(x-2)
Mrac [35]
2/3 (x+2) -1/3 (x-2)
= 2/3*x +2/3*2 -1/3x -1/3(-2) (distributive property)
= 2/3x + 4/3 -1/3x +2/3
= (2/3x -1/3x) + (4/3 + 2/3)
= 1/3x + 2

The final answer is 1/3x+2.

Hope this helps~
5 0
3 years ago
Read 2 more answers
A​ state's recidivism rate is 17​%. This means about 17​% of released prisoners end up back in prison​ (within three​ years). Su
Bezzdna [24]

Answer:

a) 2.9%

b) Option B is correct.

The prisoners must be independent with regard to recidivism.

Step-by-step explanation:

Probability that one prisoner goes back to prison = 17% = 0.17

a) The probability that two prisoners released both go back to prison = 0.17 × 0.17 = 0.0289 = 2.89% = 2.9% to 1 d.p

b) The only assumption taken during the calculation is that probability of one of the prisoners going back to prison has no effect whatsoever in the probability that another prisoner goes back to prison. That is the probability that theses two events occur are totally independent of each other.

If they weren't, we wouldn't be able to use 0.17 as the probability that the other prisoner goes back to prison too.

7 0
3 years ago
Need help now plzzz
ZanzabumX [31]

Answer: 504 in^2

Step-by-step explanation:

All you have to do is multiply all the numbers together.

3 0
3 years ago
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Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)
horsena [70]

Answer:

a.

\mathbf{r_1 = (t,0)  \implies  t = 0 \ to \ 1}

\mathbf{r_2 = (2-t,t-1)  \implies  t = 1 \ to \ 2}

\mathbf{r_3 = (0,3-t)  \implies  t = 2 \ to \ 3}

b.

\mathbf{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}

Step-by-step explanation:

Given that:

C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.

r(t) = { 0

If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),

Then:

C_1 = (0,0) \\ \\  C_2 = (1,0) \\ \\ C_3 = (0,1)

Also:

\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }

\mathbf{r_1 = (t,0)  \implies  t = 0 \ to \ 1}

\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }

\mathbf{r_2 = (2-t,t-1)  \implies  t = 1 \ to \ 2}

\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }

\mathbf{r_3 = (0,3-t)  \implies  t = 2 \ to \ 3}

b Evaluate :

Integral of (x+2y^1/2)ds

\mathtt{\int  \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \int  \limits ^1_{0} \ (t + 0)  \sqrt{1} } \\ \\ \mathtt{  \int  \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \begin {pmatrix} \dfrac{t^2}{2} \end {pmatrix} }^1_0 \\ \\  \mathtt{\int  \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \dfrac{1}{2}}

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds = \int  \limits (x+2 \sqrt{y} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 \ dt } }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds = \int  \limits 2- t + 2\sqrt{t-1}  \ \sqrt{1+1}  }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2} \int  \limits^2_1  2- t + 2\sqrt{t-1} \ dt }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} 2t - \dfrac{t^2}{2}+ \dfrac{2(t-1)^{3/2}}{3} (2)  \end {pmatrix} ^2_1}

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} 2 -\dfrac{1}{2} (4-1)+\dfrac{4}{3} (1)^{3/2} -0 \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =   \dfrac{ \sqrt{2}  }{6} \  (11 )}

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =   \dfrac{ 11 \sqrt{2}  }{6}}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds =  \int  \limits ^3_2 0+2 \sqrt{3-t}   \ \sqrt{0+1} }

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds =  \int  \limits ^3_2 2 \sqrt{3-t}   \ dt}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds =  \int  \limits^3_2 \begin {pmatrix}  \dfrac{-2(3-t)^{3/2}}{3} (2) \end {pmatrix}^3_2 }

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}

\mathtt{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}

\mathtt{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}

\mathbf{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}

5 0
2 years ago
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