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2 years ago

Show that the line 3y = 7 + 5x is perpendicular to the line 5y = 4 - 3x.

Mathematics

2 answers:

ZanzabumX [31]2 years ago

8 0

Answer:

Proven Below

Step-by-step explanation:

If we rearrange these equations, we get that 3y = 5x + 7 and that 5y = -3x + 4. Next, to find what y equals on both equations, we divide by 3 on both sides of the first equation, where we would get that y = 5/3x + 7/3, and then divide by 5 on both sides of the second equation, where we would get that y = -3/5x + 4/5. Now, to see if two lines are perpendicular to each other, we have to see if the slopes (The number being multiplied by x. Ex.) In the equation y = 3/4x + 4, the slope is 3/4) are opposite reciprocals (If you multiply them and they are equal to -1. Ex.) 3/4 * -4/3 is equal to -1, so they are opposite reciprocals). In this case, we have -3/5 and 5/3 as slopes, and -3/5 * 5/3 is equal to -1, so they are perpendicular.

saveliy_v [14]2 years ago

5 0

Answer:

The slope of the first line is 5/3

The slope for the second one is -3/5

If you multiply the 2 slope you have to end up with - 1

So 5/3*(-3/5)=-15/15 which is - 1

So yes the lines are perpenducular

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3x/2 - 2 > 7

KiRa [710]

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3 years ago

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Murrr4er [49]

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3 years ago

Suppose that θ is an acute angle of a right triangle and that sec(θ)=52. Find cos(θ) and csc(θ).

insens350 [35]

Answer:

$\cos{\theta} = \dfrac{1}{52}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

Step-by-step explanation:

To solve this question we're going to use trigonometric identities and good ol' Pythagoras theorem.

a) Firstly, sec(θ)=52. we're gonna convert this to cos(θ) using:

$\sec{\theta} = \dfrac{1}{\cos{\theta}}$

we can substitute the value of sec(θ) in this equation:

$52 = \dfrac{1}{\cos{\theta}}$

and solve for for cos(θ)

$\cos{\theta} = \dfrac{1}{52}$

side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by $\theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ$

b) since right triangle is mentioned in the question. We can use:

$\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}$

we know the value of cos(θ)=1\52. and by comparing the two. we can say that:

length of the adjacent side = 1
length of the hypotenuse = 52

we can find the third side using the Pythagoras theorem.

$(\text{hyp})^2=(\text{adj})^2+(\text{opp})^2$

$(52)^2=(1)^2+(\text{opp})^2$

$\text{opp}=\sqrt{(52)^2-1}$

$\text{opp}=\sqrt{2703}$

length of the opposite side = √(2703) ≈ 51.9904

we can find the sin(θ) using this side:

$\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}$

$\sin{\theta} = \dfrac{\sqrt{2703}}{52}}$

and since $\csc{\theta} = \dfrac{1}{\sin{\theta}}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

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3 years ago

Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the

Strike441 [17]

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

$y=Ce^{kt}$ . We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is $\frac{dy}{dt}$ . Thus, the change in the concentration of salt is found in

$\frac{dy}{dt}=$ inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

$3(\frac{y}{400})$