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Anvisha [2.4K]
3 years ago
6

Numberr 6 please help meee

Mathematics
1 answer:
expeople1 [14]3 years ago
4 0

The soda makes up 2/7 of the recipe. 2/7 of 56 is 16

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What is the answer to this
GenaCL600 [577]

30+2+0.0+0.04+0.003=32.043

8 0
3 years ago
Select the statement that does not correctly use probability to describe an event
masha68 [24]

Answer:

a

Step-by-step explanation:

Probabilty usally ends in a %

if probability of a coin landing on heads is 50% and if and

a doesnt describe probability which is why a should be the anwser

though d says 120% the person running the race has full convidence describing themselves that they will win

a doesnt describe any probability it just says 1!! but thats why a is the anwser because its inncorrect!

5 0
3 years ago
4<br> 74<br> Solve the inequality 2(4x + 1) &lt;3(2x - 3)
Katena32 [7]

Step-by-step explanation:

2(4x + 1) < 3(2x - 3)

8x + 2 < 6x - 9

8x - 6x < -9 -2

2x < -11

x < -11/2

8 0
2 years ago
For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f
Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0

We want to find f such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)

\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C

\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}

so \vec F is conservative.

2.

\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)

\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)

\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C

\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}

so \vec F is conservative.

3.

\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0

so \vec F is not conservative.

4.

\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)

\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)

\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C

\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}

so \vec F is conservative.

4 0
3 years ago
Four friends paid a total of 32 dollars for movie tickets. What is the ratio 32 dollars for4 people written as a unit rate
coldgirl [10]
8 is the ratio for the unit rate
3 0
3 years ago
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