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belka [17]
3 years ago
5

What quadratic equation has roots 7 and -2/3

Mathematics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

(x-7)(x+2/3)

Step-by-step explanation:

To easily find an equation with said roots, use the intercept form:

y=a(x-p)(x-q), where p and q are your roots.

y=(x-7)(x-(-2/3) = y=(x-7)(x+2/3)

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The measurements 45.367 cm and 43.43 cm are made in the same unit of measure. Which measurement is less precise? Explain your an
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Step-by-step explanation:

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2 years ago
Use the Fundamental Theorem of Line Integrals to calculate ∫c F⃗ ⋅dr⃗ exactly, if F⃗ =3x2/3i⃗ +ey/5j⃗ , and C is the quarter of
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It looks like the vector field is

<em>F</em><em>(x, y)</em> = 3<em>x</em> ^(2/3) <em>i</em> + <em>e</em> ^(<em>y</em>/5) <em>j</em>

<em></em>

Find a scalar function <em>f</em> such that grad <em>f</em> = <em>F</em> :

∂<em>f</em>/∂<em>x</em> = 3<em>x</em> ^(2/3)   =>   <em>f(x, y)</em> = 9/5 <em>x</em> ^(5/3) + <em>g(y)</em>

=>   ∂<em>f</em>/∂<em>y</em> = <em>e</em> ^(<em>y</em>/5) = d<em>g</em>/d<em>y</em>   =>   <em>g(y)</em> = 5<em>e</em> ^(<em>y</em>/5) + <em>K</em>

=>   <em>f(x, y)</em> = 9/5 <em>x</em> ^(5/3) + 5<em>e</em> ^(<em>y</em>/5) + <em>K</em>

(where <em>K</em> is an arbitrary constant)

By the fundamental theorem, the integral of <em>F</em> over the given path is

∫<em>c</em> <em>F</em> • d<em>r</em> = <em>f</em> (0, 1) - <em>f</em> (1, 0) = 5<em>e</em> ^(1/5) - 34/5

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