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3 years ago

12

I'm having a hard time with this. A new housing development extends 4 miles in one direction, makes a right turn, and then con-

tinues for 3 miles. A new road runs between the beginning and ending points of the development. What is the perimeter of the triangle formed by the homes and the road? What is the area of the housing development?

1 answer:

Mademuasel [1]3 years ago

8 0

Answer:

perimeter = 12 miles

area = 6 square miles

Step-by-step explanation:

Since it makes a right triangle, use the Pythagorean Formula.

3^2+4^2=c^2

9+16=c^2

25=c^2

5=c, so the hypotenuse of the right triangle is 5.

Perimeter = 3+4+5 = 12 miles

area = 1/2bh (1/2 base times height)

=1/2x3x4

=6

Area = 6 square miles

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Natali5045456 [20]

Divide 500 by 4

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4 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:

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

3 years ago

A number k is multiplied by 3/4. the product is 12. What is the value of k?

elena55 [62]

Answer:

3/4 × 12/1

3×12/4×1

36/4

9

4 0

3 years ago

Read 2 more answers

What expression is equivalent to ^3 square root of 2y^3 times 7 square root of 18y

Mumz [18]

Answer:

$\sqrt[3]{2y^3} * 7\sqrt{18y} = 21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

Step-by-step explanation:

The question is poorly formatted.

Given

$\sqrt[3]{2y^3} * 7\sqrt{18y}$

Required

Derive an equivalent expression

$\sqrt[3]{2y^3} * 7\sqrt{18y}$

Express 18 as 9 * 2

$\sqrt[3]{2y^3} * 7\sqrt{9 * 2y}$

Split the expression as follows:

$\sqrt[3]{2y^3} * 7\sqrt{9} * \sqrt{2y}$

Take positive square root of 9

$\sqrt[3]{2y^3} * 7*3 * \sqrt{2y}$

$\sqrt[3]{2y^3} * 21 * \sqrt{2y}$

$21*\sqrt[3]{2y^3} * \sqrt{2y}$

The cube root can be rewritten to give:

$21*\sqrt[3]{2}*\sqrt[3]{y^3} * \sqrt{2y}$

$\sqrt[3]{y^3} = y^{3*\frac{1}{3}} = y$

So, we have:

$21*\sqrt[3]{2} * y * \sqrt{2y}$

Rewrite as:

$21y *\sqrt[3]{2} * \sqrt{2y}$

Split $\sqrt{2y$

$21y *\sqrt[3]{2} * \sqrt{2} * \sqrt{y}$

Collect Like Terms

$21y*\sqrt{y} *\sqrt[3]{2} * \sqrt{2}$

Represent in index form

$21y*y^{\frac{1}{2}} *2^\frac{1}{3} *2^\frac{1}{2}$

Apply law of indices

$21*y^{1+\frac{1}{2}} *2^{\frac{1}{3} +\frac{1}{2} }$

$21*y^{\frac{2+1}{2}} *2^{\frac{2+3}{6}}$

$21*y^{\frac{3}{2}} *2^{\frac{5}{6}}$

$21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

Hence:

$\sqrt[3]{2y^3} * 7\sqrt{18y} = 21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

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