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2 years ago

Nine divide by the quantity x plus y

Mathematics

1 answer:

Vadim26 [7]2 years ago

5 0

7(x + 9 ) - x

Explanation:

the quantity x plus 9 is written x + 9 , plus meaning add.

times the quantity means multiply the quantity x + 9 multiplied by 7

this is written 7(x + 9 )

finally minus x means subtract x

putting it all 'together' we get 7(x + 9 ) - x

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Find three consecutive integers such that three times The largest increased by two is equal to five times the smallest increased

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3 years ago

Choose the graph which represents the solution to the inequality. 12x + 4 ≤ -8

user100 [1]

Answer:

Step-by-step explanation:

Steps:

$12x+4-4\leq -8$

Subtract 4 from both sides:

$12x+4-4\leq -8-4$

Simplify:

$12x\leq \frac{-12}{12}$

Divide both sides by 12:

$\frac{12x}{12} \leq \frac{-12}{12}$

Simplify:

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3 years ago

Which numbers are a distance of 2 units from 8 on a number line?

Aleonysh [2.5K]

6 and 10 are 2 units away from a positive 8 on the number line

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3 years ago

Read 2 more answers

Which expression is NOT equivalent to (a + b)(x + y)? A) (a + x)(b + y) B) (b + a)( y + x) C) (a + b)x + (a + b)y D) a(x + y) +

Tanzania [10]

Answer:

Step-by-step explanation:

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2 years ago

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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2 years ago