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3 years ago

Solve for Z. 42=-7(z-3)

Mathematics

2 answers:

crimeas [40]3 years ago

6 0

42= -7(z-3) distribute in parenthesis
42= -7z +21 isolate the variable z
7z=21-42
7z= -21
Z= -21/7
Z= -3

elixir [45]3 years ago

3 0

42/7=Z-3

6=Z-3

Z=6+3

Z=9

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a. $\mathbf{36 \sqrt{5}}$

b. $\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

Step-by-step explanation:

Evaluate integral _C x ds where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

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where;

x = t , y = t/2

the derivative of x with respect to t is:

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the derivative of y with respect to t is:

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and t varies from 0 to 12.

we all know that:

$ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt$

∴

$\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt$

$= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt$

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b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t ; y = 3t²

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt} = 6t$

$ds = \sqrt{1+36 \ t^2} \ dt$

Hence; the integral _C x ds is:

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

Let consider u to be equal to 1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

$tdt = \dfrac{du}{76}$

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

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$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

$\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}$

$= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}$

$\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}$

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