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

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

Read 2 more answers

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