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4 years ago

Solve for x. 4x - 4 < 8 and 9x + 5 > 23

Mathematics

1 answer:

nasty-shy [4]4 years ago

7 0

Answer:

$x$ and $x>2$

Step-by-step explanation:

$4x - 4 < 8\\4x$

$9x+5>23\\9x>23-5\\9x>18\\x>\frac{18}{9}\\ x>2$

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I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT

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Answer:

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Step-by-step explanation:

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3 years ago

How do I turn this into slope-intercept form

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3 years ago

Cathy wants to pack 100 gifts. The packaging includes a closed box made out of cardboard. The dimensions of the box are 11cm Lon

wlad13 [49]

Answer:

$10800\ \text{cm}$

Step-by-step explanation:

l = Length = 11 cm

w = Width = 8 cm

h = Height = 8 cm

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3 years ago

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

horsena [70]

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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3 years ago

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