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3 years ago

5

Ms. Smiths block 2 class has 30 students. 2/3 of them are absent on a field trip for science. How many students in the class are

absent because they are on the field trip?

2 answers:

Olenka [21]3 years ago

7 0

20 because if you think about it 2/3 is equal to 20/30.

Papessa [141]3 years ago

5 0

20 of the kids are absent

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Help please due tonight <br><br>50 points + brainliest

Leviafan [203]

Answer:

7. -37/36 or -1 1/36

8. -3/10

9. 1/24

10. -118/21 or -5 13/21

11. 97/12 or 8 1/12

12. 49/8 or 6 1/8

Step-by-step explanation:

To subtract fractions, they must have the same denominator. The common denominator will be the LCM (lowest common multiple) of the two denominators. Then, combine the fractions and simplify.

7. $-\frac{1}{4}-\frac{7}{9}$ Find a common denominator

$=-\frac{1*9}{4*9}-\frac{7*4}{9*4}$

$=-\frac{9}{36}-\frac{28}{36}$

$=\frac{-9-28}{36}$ Combine the fractions

$=\frac{-37}{36}$ Answer. Already in lowest terms.

$=-1\frac{1}{36}$ Keep a negative number. 37 goes into 36 one time, leaving 1 remainder (becomes numerator)

8. $\frac{3}{5}-\frac{9}{10}$ Find a common denominator

$=\frac{3*2}{5*2}-\frac{9}{10}$

$=\frac{6}{10}-\frac{9}{10}$

$=\frac{6-9}{10}$ Combine the fractions

$=\frac{-3}{10}$ Answer in simplest form.

9. $\frac{5}{8}-\frac{7}{12}$ Find common denominator

$=\frac{5*3}{8*3}-\frac{7*2}{12*2}$

$=\frac{15}{24}-\frac{14}{24}$

$=\frac{15-14}{24}$ Combine the fractions

$=\frac{1}{24}$ Answer in simplest form

10. $-1\frac{1}{3}-4\frac{2}{7}$ Translate to improper fraction

$=-\frac{4}{3}-\frac{30}{7}$ Multiplied each whole number with base, added to numerator

$=-\frac{28}{21}-\frac{90}{21}$ Found common denominator

$=\frac{-28-90}{21}$ Combined fractions

$=\frac{-118}{21}$ Answer in improper form

$=-5\frac{13}{21}$ Answer as a mixed number

11. $5\frac{5}{6}-(-2\frac{1}{4})$ Convert to improper fraction

$\frac{35}{6}-(-\frac{9}{4})$ Two minuses make a plus

$=\frac{35}{6}+\frac{9}{4}$ Find common denominator

$=\frac{35*2}{6*2}+\frac{9*3}{4*3}$

$=\frac{70}{12}+\frac{27}{12}$

$=\frac{70+27}{12}$ Combined fractions

$=\frac{97}{12}$ Answer as improper fraction

$=8\frac{1}{12}$ Answer as mixed fraction

12. $12\frac{1}{2}-6\frac{3}{8}$ Convert to improper fraction

$\frac{25}{2}-\frac{51}{8}$

$\frac{25*4}{2*4}-\frac{51}{8}$ Find common denominator

$\frac{100}{8}-\frac{51}{8}$

$\frac{100-51}{8}$ Combined fractions

$\frac{49}{8}$ Answer as improper fraction

$8\frac{1}{8}$ Answer as mixed number

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Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 2yi + xzj + (x

Lady_Fox [76]

By Stokes' theorem, the line integral of $\vec F$ over $C$ is given by the surface integral of the curl of $\vec F$ over $S$ , where $S$ is the region of intersection of the plane $z=y+6$ and the cylinder $x^2+y^2=1$ with $S$ having positive/upward orientation.

Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k$

with $0\le u\le1$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k$

The curl of $\vec F$ is

$\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k$

Then the line integral is equivalent to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(5u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{5\pi}$

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