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3 years ago

Write PQ−⇀− with P(2,−4) and Q(3,6) in component form.

Mathematics

1 answer:

Doss [256]3 years ago

7 0

Answer:

1,10

Step-by-step explanation:

The initial point of the vector is P(2,−4) and the terminal point of the vector is Q(3,6).

Subtract the coordinates of the initial point from the coordinates of the terminal point.

PQ−⇀−=<xQ−xP,yQ−yP>

Substitute the coordinates of the given points.

PQ−⇀−=<3−2,6−(−4)>

Simplify.

PQ−⇀−=<1,10>

Therefore, the component form of vector PQ−⇀− is <1,10>.

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Range of f(x)=-x+5 with domain of [-3,2] in the same notation as the domain.

Oksi-84 [34.3K]

The range would be [8, 3]

In order to find the range, simply plug each end of the domain into the equation.

f(x) = -x + 5

f(-3) = -(-3) + 5

f(-3) = 3 + 5

f(-3) = 8

Then the other side

f(x) = -x + 5

f(2) = -2 + 5

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3 years ago

Jamie throws a ball into the air with an initial upward velocity of 68 feet per second from a height of 12 feet. Write a quadrat

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Answer:

$h(t)=68t+12$

Step-by-step explanation:

let t= time in seconds, h=height in ft

height at $t_0=12ft$ = $h_o$

Velocity,v= $68m/s$

Height= $velocity*time=vt$

$h_t=h_0+h_(t-0)$

Therefore height after $t_n$ is obtained by adding height gained to initial height.

$h(t)=68t+12$

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4 years ago

1. Write the angles in order from smallest to largest 45 40.5 R P 57

algol13

Hello!! :D

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3 years ago

What statistics are needed to draw a box plot?

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4 years ago

In each problem 7 through 14, verify that each given function is a solution of the differential equation. Y" - y = 0; y1(t) = et

Lilit [14]

Answer:

Since L.H.S = R.H.S = 0, for both $y_{1} (t) = e^{t}$ and $y_{2} (t) = cosh(t)$ , y₁ and y₂ both satisfy the equation y" - y = 0 and are thus solutions to the differential equation.

Step-by-step explanation:

To check whether the given functions are solutions the given differential equation, we differentiate the functions and then insert it into the given equation.

So y" - y = 0 and

$y_{1} (t) = e^{t}\\y_{1}' (t) = e^{t}\\ y_{1}" (t) = e^{t}$