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Gnesinka [82]
3 years ago
9

SOMEBODY PLEASE HELP

Mathematics
2 answers:
Ivanshal [37]3 years ago
7 0

AB---------DE

BC---------EF

AC----------DF

uranmaximum [27]3 years ago
4 0
AB corresponds to DE
BC corresponds to EF
AC corresponds to DF
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A person 2.1 m tall casts a shadow 5.3 m long. At the same time, a building casts a shadow 34
bulgar [2K]
Height. Shadow
2.1 m. 5.3 m
X m. 34 m
X=2.1•34/5.3
X=71.4/5.3
X=13.47 m
X~13.5 m
4 0
3 years ago
((PLEASE ANSWER WITH A B C or D))
natali 33 [55]

Answer:

  D.  60°

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that the relationship of interest is ...

  Sin = Opposite/Hypotenuse

  sin(M) = ON/OM = 4(√3)/8 = (√3)/2

  M = sin⁻¹((√3)/2) = 60°

8 0
3 years ago
The measure of angle A is 15°, and the length of side BC is 8. What are the lengths of the other two sides, rounded to the neare
GaryK [48]
AC=29.9
AB=30.9 that is the answer for this question
3 0
3 years ago
Read 2 more answers
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DochEvi [55]
Angle 2 and angle 11 are alternate exterior angles. Line G and line L are parallel lines. Think of them as train tracks. On the outside or exterior of the train tracks is angle 2 and angle 11.

Angle 2 is on the right side of the transversal line, while angle 11 is on the left side of the transversal line.

So this is why they are alternate exterior angles. Because line G and line L are parallel lines, this means the alternate exterior angles are congruent (by the alternate exterior angle theorem). 

Since angle 2 is 115 degrees, angle 11 must also be 115 degrees

Answer: C) 115
7 0
3 years ago
Use Stokes' theorem to evaluate ∬S(∇×F)⋅dS where F(x,y,z)=−3yzi+3xzj+14(x2+y2)zk and S is the part of the paraboloid z=x2+y2 tha
Usimov [2.4K]

Stokes' theorem says the integral of the curl of \vec F over S is equal to the integral of \vec F along the boundary of S, with counterclockwise orientation (when viewed from above). This boundary is the circle x^2+y^2=1 set in the plane z=1.

Parameterize this path by

\vec r(t)=\cos t\,\vec\imath+\sin t\,\vec\jmath+\vec k

with 0\le t\le2\pi. Then

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_{\partial S}\vec F\cdot\mathrm d\vec r

=\displaystyle\int_0^{2\pi}(-3\sin t\,\vec\imath+3\cos t\,\vec\jmath+14\,\vec k)\cdot(-\sin t\,\vec\imath+\cos t\,\vec\jmath)\,\mathrm dt

=\displaystyle3\int_0^{2\pi}\mathrm dt=\boxed{6\pi}

4 0
3 years ago
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