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2 years ago

Combine like terms to find the equivalent expression of 5 y minus 3 y + 12.

Mathematics

2 answers:

satela [25.4K]2 years ago

7 0

Answer:

2y + 12

Step-by-step explanation:

Combine like terms. The terms are numbers with a y, & then numbers without.

So combine the y's, leave the 12.

5y - 3y + 12

5 - 3 = 2

2y + 12

Hope this helps. :0)

Scorpion4ik [409]2 years ago

7 0

Answer: correct me if im wrong but 2y + 12?

Step-by-step explanation:

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In the Itty Bitty High School, there are 85 students. There are 27 students who take French, 51 who take Geometry and 38 who tak

andrew-mc [135]

Answer:

58 students

Step-by-step explanation:

Given

$Total = 85$

$French = 27$

$Geometry = 51$

$History = 38$

$None = 5$

$All = 2$

Required

Students that take only 1 subjects

The attached image illustrates the question.

For French only (x), we have:

$x + 7 -2+10-2+2=27$

$x + 15=27$

Collect like terms

$x =- 15+27$

$x =12$

For Geometry only (y), we have:

$y+10-2+15-2+2=51$

$y+23=51$

Collect like terms

$y=-23+51$

$y=28$

For History only (z), we have:

$z +15-2+7-2+2=38$

$z +20=38$

Collect like terms

$z =-20+38$

$z =18$

Students with one subject is then calculated as:

$1\ subject = x+y+z$

$1\ subject = 12 + 28 + 18$

$1\ subject = 58$

5 0

2 years ago

A pair of vertical angles have measures (2y+5)° and (4y)°.

Molodets [167]

I believe with vertical angles, they are equal to each other.

Therefore,

2y + 5 = 4y

subtract 2y from both sides

5 = 2y

divide 2 from both sides

y = 5/2

8 0

3 years ago

Read 2 more answers

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.

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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

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VashaNatasha [74]

Answer:

Step-by-step explanation:

As it is a list of prime number, the sequence continues with 29 being the next prime number after 23.

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2 years ago

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Dmitry [639]

Answer:

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