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fenix001 [56]
3 years ago
7

Kevin says the line P and m will eventually Intersect is Kevin correct

Mathematics
1 answer:
bonufazy [111]3 years ago
7 0

no because its (they are) parallel

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two airplane fly toward each other from points 900 miles apart, at the rate, per hour, of 300 miles and 250 miles. when will the
mojhsa [17]
They complete a 900-mile distance with their combined speed of (300+250=550) mph
Therefore, they meet after 900m/550mph=18/11h=1.6364h
6 0
3 years ago
Complete the equation of the line through (-6,-5)(−6,−5)left parenthesis, minus, 6, comma, minus, 5, right parenthesis and (-4,-
Alex

Answer:

y = 1/2 x -2

Step-by-step explanation:

rise/run from (-6, -5) to (-4, -4) = 1/2

(rise 1, run 2)

meaning that slope (m) = 1/2

we can now rewrite our points in y=mx + b formatting:

y = mx + b

-5 = (1/2)(-6) + b

-4 = (1/2)(-4)+b

now, let's solve for b

-5 = (1/2)(-6) + b

-5 = -3 + b

-2 = b

-4 = (1/2)(-4)+b

-4 = -2 + b

-2 = b

So, the equation for our line is y = \frac{1}{2} x- 2

hope this helps!! have a lovely day :)

4 0
2 years ago
Read 2 more answers
Given: segment AB || segment DE, C is the midpoint of segment DB. Prove: ΔACB ≅ ΔECD Fill in the missing reason for the proof. A
OleMash [197]

Answer:

Given: Segment AB || segment DE, C is the midpoint of segment DB.

Prove: ΔA CB ≅ ΔE CD

Proof: In ΔA CB and ΔE CD

 C is the Mid point of B D.

BC=C D→ definition of midpoint

∠A CB= ∠ EC D→→vertical angles are congruent

∠BAC=∠DEC→→[AB║DE,so alternate angles are equal]

→→ΔA CB ≅ ΔE CD[A AS or A SA]

Option B: vertical angles are congruent

4 0
3 years ago
Read 2 more answers
There are 12 third grade students and 15 fourth grade students in the band. They will travel to a convert in cars that can hold
Novay_Z [31]
The answer is the first one (12+15) / 4
6 0
3 years ago
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evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.
meriva

We have

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt

and

\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k

\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k

so the line integral is equilvalent to

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}

\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}

7 0
2 years ago
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