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inna [77]
3 years ago
9

Write the point-slope form of the equation of the line through the point (3,-5) with a slope of -4/3.

Mathematics
1 answer:
hodyreva [135]3 years ago
4 0

Answer:

asda

Step-by-step explanation:

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Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl
GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3

Substitute the value of \lambda in the parametric equations:

x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

7 0
3 years ago
Write 25/20 as a percentage
kolezko [41]
125% my guys im sorry if its wrong :(
5 0
2 years ago
Read 2 more answers
What is 4-(-37)+(8)-(34) Pls solve brainlliest
Anvisha [2.4K]

Answer:

15

Step-by-step explanation:

First we need to remove the parenthesis

4 + 37 + 8 - (34)

4 + 37 + 8 -34

Add 4 and 37

41 + 8 - 34

Add 41 and 8

49 - 34

Subtract

15

6 0
2 years ago
What is the answer to this problem and how do you solve the problem? <br><br><br> 2x^-2 y^-3/(3a^-4)
Damm [24]
Well first do you parentheses and the multiply all of your exponets 
3 0
3 years ago
A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,
algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}

Where v_i is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is [0, \pi/2]. The critical  points of the function are those who make d'(\theta)=0:

d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}

d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...

The critical value inside the interval is \pi/4.

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft

The second step is to find the values of the function at the endpoints of the interval:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft

The biggest value of f is gived by \pi/4, therefore \pi/4 is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

4 0
3 years ago
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