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3 years ago

How to do this question plz answer me

Mathematics

1 answer:

Sati [7]3 years ago

7 0

Answer:

$d = \frac{4 - 3c}{c + 1}$

Step-by-step explanation:

To make d the subject of formula, we need to rearrange the equation such that we arrive at d= _____.

$c = \frac{4 - d}{d + 3}$

Remove the fraction by multiplying (d +3) on both sides:

$c(d + 3) = 4 - d$

Expand:

$cd + 3c = 4 - d$

Bring all the d terms to one side and move the others to the other side of the equation:

$cd + d = 4 - 3c$

Factorise d out:

$d(c + 1) = 4 - 3c$

Divide by (c +1) on both sides:

$d = \frac{4 - 3c}{c + 1}$

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In her piggy bank, caitlin has 5.24 dollars the sum is made up of an equal number of four coins from 1 cent, 5 cents, 10 cents,

Ludmilka [50]

Answer:

1c, 5c, 25c and 1 dollar

Step-by-step explanation:

With 4 coins of 1c : 4 cents

4 coins of 5c: 20 cents

4 coins of 25c: 1 dollar

4 coins of 1 dollar: 4 dollars

So, 4c + 20c + 1 dollar + 4 dollars = 5.24 dollars

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3 years ago

What is the range of f(x) = 4x?

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4 years ago

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What is the value of the expression 30+(-6)? A -180 B -5 C 5 D 24

Evgesh-ka [11]

Answer:

Step-by-step explanation:

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30+(-6) =

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2 years ago

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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

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3 years ago

What the heck am I doing something something something I’m doing this. So there’s enough writing.

AleksandrR [38]

Answer:

are you okay ? that didnt really make sense sorry love

Step-by-step explanation:

4 0

3 years ago

How to do this question plz answer me ​

How to do this question plz answer me