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GarryVolchara [31]
2 years ago
15

Set R contains all integers from 10 to 125, inclusive, and Set T contains all integers from 82 to 174, inclusive. How many integ

ers are included in R, but not in T?
Mathematics
1 answer:
SCORPION-xisa [38]2 years ago
3 0

Answer: PLEASE GIVE BRAINLIEST

72 integers

Step-by-step explanation:

Starting From 82-174 every integer is included in T so every integer that is not included in T but included in R would be 82-10 so every integer from 10 to 82.

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EXAMPLE 5 If F(x, y, z) = 4y2i + (8xy + 4e4z)j + 16ye4zk, find a function f such that ∇f = F. SOLUTION If there is such a functi
Valentin [98]

If there is such a scalar function <em>f</em>, then

\dfrac{\partial f}{\partial x}=4y^2

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}

\dfrac{\partial f}{\partial z}=16ye^{4z}

Integrate both sides of the first equation with respect to <em>x</em> :

f(x,y,z)=4xy^2+g(y,z)

Differentiate both sides with respect to <em>y</em> :

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}=8xy+\dfrac{\partial g}{\partial y}

\implies\dfrac{\partial g}{\partial y}=4e^{4z}

Integrate both sides with respect to <em>y</em> :

g(y,z)=4ye^{4z}+h(z)

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :

f(x,y,z)=4xy^2+4ye^{4z}+h(z)

\dfrac{\partial f}{\partial z}=16ye^{4z}=16ye^{4z}+\dfrac{\mathrm dh}{\mathrm dz}

\implies\dfrac{\mathrm dh}{\mathrm dz}=0

Integrate both sides with respect to <em>z</em> :

h(z)=C

So we end up with

\boxed{f(x,y,z)=4xy^2+4ye^{4z}+C}

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Step-by-step explanation:

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For which discriminant is the graph possible<br> b2-4ac=0<br><br> b2-4ac=-1<br><br> b2-4ac=4
VARVARA [1.3K]

Answer:

The graph is possible for b^2-4ac=4

Step-by-step explanation:

we know that

The discriminant of a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

D=b^2-4ac

If D=0 the quadratic equation has only one real solution

If D>0 the quadratic equation has two real solutions

If D<0 the quadratic equation has no real solution (complex solutions)

In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)

so

b^2-4ac > 0

therefore

The graph is possible for b^2-4ac=4

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