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fiasKO [112]
3 years ago
13

2

Mathematics
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:?

Step-by-step explanation:

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HELP ASAP!!!!!!! NO LINKS OR FAKE ANSWERS OR I WILL REPORT U!
Semmy [17]

Answer:

Step-by-step explanation:

54, 1\frac{2}{3} , 1, 0.5, 0.03, 0, - 1, - 1\frac{3}{4} , - 4, - 103

5 0
3 years ago
Your friend claims that it is possible to draw a right triangle where the cosine of either angle (theta) is exactly the same val
avanturin [10]

Answer:

B. No

Cos \ \theta\neq Cos (90-\theta)\textdegree

Step-by-step explanation:

-A right angle triangle has two complimentary acute angles and one right angle.

-\theta is usually one of the acute angles and is equivalent to 90º minus it's complimentary acute angle.

-Complimentary angles add up to 90º.

#For complimentary angles:

Sin \ \theta=Cos \ (90-\theta)\textdegree\\\\Cos \ \theta=Sin(90-\theta)\textdegree\\\\\therefore Cos \ \theta\neq Cos (90-\theta)\textdegree

The two acute angles cannot have the same Cosine value.

Hence, she's not correct.

5 0
3 years ago
Suppose an airplane climbs 15 feet for every 40 feet it moves forward. What is the slope of this airplanes ascent?
Goryan [66]

Answer:

3/8

Step-by-step explanation:

Slope is the same as rise over run, or y/x. In this case, rise over run is 15 over 40, or 15/40. Simplified, this is 3/8.

8 0
3 years ago
Asappppp!!!!!!!!!!!!!!!!!!!!!!!!
Scrat [10]
42 because x is equal to its corresponding side
7 0
4 years ago
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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