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Gemiola [76]
3 years ago
14

2/3 divided by 1/4 in its simplest form

Mathematics
2 answers:
rewona [7]3 years ago
5 0

Answer:

8/3

Step-by-step explanation:

to divide by a fraction multiply the reciprocal

2/3×4

calculate the product

2×4=8

8/3

Sveta_85 [38]3 years ago
4 0
Answer

2 2/3

2/3 divided by 1/4

Swap the 1/4 making it 4/1
Then times the fractions together
Giving you
8/3
Simplify

= 2 2/3
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2 years ago
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Write the equation for a line in standard form (Ax+By+C) that is perpendicular to y = 3x -
salantis [7]

Answer:

x+3y-6=0

Step-by-step explanation:

given eqn is y=3x-2 which is 3x-y-2=0

the eqn of line perpendicular to given eqn is -x+3y+k=0

it passes through (6,4)

-6+3*4+k=0

or,. -6+12+k=0

or, k= -6

therefore, the eqn of line perpendicular to given eqn is x+3y-6=0

8 0
3 years ago
If the point (4,-2) is included in a direct variation relationship, which point also belongs in this direct variation? A. (-4,2)
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Answer:

(-4,2) belongs in this direct variation.

Step-by-step explanation:

Let the direct variation relationship is expressed by the equation y = mx ........ (1), where x and y are in direct variation and k is the variation constant.

Now, the point (4,-2) is included in the direct variation relationship, then from equation (1) we get, -2 = 4m

⇒ m =  - \frac{1}{2}

Therefore, the equation (1) becomes y = - \frac{1}{2}x

⇒ x + 2y = 0 ......... (2)

Now, from the given four options only the point (-4,2) satisfies the relation (2).

Hence, (-4,2) belongs in this direct variation. (Answer)

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Determine whether the given differential equation is exact. If it is exact, solve it. (If it is not exact, enter NOT.) (tan(x) −
never [62]

Answer:

f(x,y)=ln secx+cosx siny+C

Step-by-step explanation:

We are given that DE

(tanx-sinx siny)dx+cosxcosydy=0

We have to determine given DE is exact or not.

Compare it with Mdx+Ndy=0

M=tanx-sinx siny

N=cosxcosy

\frac{\partial M}{\partial y}=M_y=-sinxcosy

\frac{\partial N}{\partial x}=N_x=-sinxcosy

Therefore, M_y=N_x

If DE is exact then M_y=N_x

Hence,it is exact.

M=\frac{\partial f}{\partial x}=tanx-sinxsiny

Integrate w.r.t x on both sides

f(x,y)=\int(tanx-sinxsiny)dx

f(x,y)=lnsecx+cosxsiny+\phi(y)...(1)

By using the formula

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Differentiate partially  equation (1) w.r.t y

\frac{\partial f}{\partial y}=cosxcosy+\phi'(y)

By using the formula:

\frac{d(sinx)}{dx}=cosx

N=\frac{\partial f}{\partial y}=cosxcosy=cosxcosy+\phi'(y)

\phi'(y)=cosxcosy-cosxcosy=0

\phi'(y)=0

Integrate  w.r.t y

\phi(y)=C

Substitute the value

f(x,y)=ln secx+cosx siny+C

7 0
3 years ago
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