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Karo-lina-s [1.5K]
3 years ago
12

HELP ME PLZZZZ I'LL GIVE ANYONE 100 FOR THE CORRECT ANSWER

Mathematics
2 answers:
Travka [436]3 years ago
3 0

Answer:

The symbol '<' fits best.

Step-by-step explanation:

18 ÷ 6 + 3 □ 6 + 12 ÷ 3

=> 3 + 3 □ 6 + 4

=> 6 □ 10

Therefore, the symbol '<' fits best.

Hoped this helped.

PilotLPTM [1.2K]3 years ago
3 0

Answer: a.<

Step-by-step explanation:

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Anton [14]

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16

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ez boi V = Bh V=16

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Which number is rational? ​
Romashka [77]

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5.3333333...

Step-by-step explanation:

5.\overline3 = \dfrac{53-5}{9} = \dfrac{48}9

4 0
2 years ago
Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks. Last
lisabon 2012 [21]

Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks....

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4 years ago
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I really hope you can see it if you can’t please ask me what it says cause I really need help
soldier1979 [14.2K]

Answer:

1.   20 cm^2

2.  40 cm^2

3.   14 cm^2

4    49 cm^2

Step-by-step explanation:

Area of triangle= 1/2(base)*(height)

Area of rectangle= (base)*(height)

1. 1/2(10)(4)=20

2. (8)(5)=40

3.1/2(4)(7)=14

4. (7)(7)=49

5 0
3 years ago
Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)
choli [55]

Answer: Hello mate!

Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

for all the elements in A, the, for all the elements on A you get:

\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}

\frac{du}{dx}  =  \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}

and now lets derivate this with respect to y.

using that \frac{d(sec(x))}{dx}= sec(x)*tan(x)

\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy}  = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)

Now if we first derivate by y, we get:

\frac{du}{dy}  =  \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}

and now we derivate by x:

\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy}  = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)

the mixed partial derivates are equal :)

7 0
3 years ago
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