Please help!!!! I need help woth the select the quadrent or axis where each order pair is located on the coorninate plane

Find the equation of the lines, that is parallel to the line 9x-8y=1, and tangent to the elipses 9x^2+16y^2=52.

lara31 [8.8K]

Answer:

The equations of the lines that satisfy these conditions are

y = (9/8)x + (13/4)

And

y = (9/8)x - (13/4)

They can be written further as

8y = 9x + 26

and

8y = 9x - 26

Step-by-step explanation:

The line parallel to the line 9x - 8y = 1 has the same slope as 9x - 8y = 1

Writing the equation of the line in the (y = mx + c) form

8y = 9x - 1

y = (9/8)x - 1

So, the line(s) we are looking for has/have slope(s) of (9/8).

Its equation is then

y = (9/8)x + k

This line is then said to be tangent to the ellipses

9x² + 16y² = 52

So, since the line is a tangent to the ellipses, it will have the same coordinates as the ellipses at the point of contact.

9x² + 16y² = 52

y = (9/8)x + k

y² = (81/64)x² + (18k/8)x + k²

y² = (81/64)x² + (9k/4)x + k²

Substituting this into the ellipses equation,

9x² + 16y² = 52

9x² + 16[(81/64)x² + (9k/4)x + k²] = 52

9x² + (81/4)x² + 36kx + 16k² = 52

29.25x² + 36kx + 16k² - 52 = 0

Taking note that the two lines that will be tangent to the ellipses and parallel to the line given, to establish tangency, we make the discriminant of this equation 0. Hence, the roots of that equation will have equal roots.

So, the discriminant, b² - 4ac = 0

29.25x² + 36kx + 16k² - 52 = 0

a = 29.25

b = 36k

c = 16k² - 52

(36k)² - (4)×(29.25)×(16k² - 52) = 0

1296k² - 1872k² + 6084 = 0

576k² = 6084

k = ±(13/4)

So, the equation(s) of the line(s) required is

y = (9/8)x ± (13/4)

So, the equations include

y = (9/8)x + (13/4)

And

y = (9/8)x - (13/4)

Multiplying by 8, we obtain

8y = 9x + 26

8y = 9x - 26

Hope this Helps!!!

6 0

3 years ago

I have 200 coins to put into 4 bags I put the coins into each bag so that each bag has 2 mote coins than the one before How many

Allushta [10]

First bag has 47 coins and second bag has 49 coins and third bag has 51 coins and fourth bag has 53 coins

Solution:

Given that,

Total number of coins = 200

Number of bags = 4

I put the coins into each bag so that each bag has 2 more coins than the one before

Therefore,

Each bag has 2 more coins than the one before. Based on this we can say,

Let "x" be the number of coins put in first bag

Then, x + 2 is the number of coins put in second bag

Then, x + 4 is the number of coins put in third bag

Then, x + 6 is the number of coins put in fourth bag

We know that,

Total number of coins = 200

$x + x + 2 + x + 4 + x + 6 = 200\\\\4x + 12 = 200\\\\4x = 200-12\\\\4x = 188\\\\x = 47$

Thus,

Coins put in first bag = x = 47

Coins put in second bag = x + 2 = 47 + 2 = 49

Coins put in third bag = x + 4 = 47 + 4 = 51

Coins put in fourth bag = x + 6 = 47 + 6 = 53

Thus number of coins in each bag are found

4 0

3 years ago

Find the Area of the Shaded Region.

agasfer [191]

Answer:

I think it should be 32

Step-by-step explanation:

11 x 7= 77

9 x 5=45

77-45=32

3 0

3 years ago

segment BC is an are of a circle with radius 30.0 cm, and point P is at the center of curvature of the arc Segment DA is an arc

Elan Coil [88]

Answer:

The magnetic field is $B = 3.87 *10^{-6} \ T$

Direction is inward

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The radius of BC is $r_{BC} = 30 \ cm = 0.30 \ m$

The radius of DA is $r _{DA} = 20 \ cm = 0.20 \ m$

The length of CD is $r_{CD} = 10 cm = 0.1 \ m$

The length of AB is $r_{AB} = 10 cm = 0.1 \ m$

The current is $I = 11.4A$

The magnetic field is mathematically represented as

$B = B_{BC} + B_{DA} + B_{CD} + B_{AB}$

$B = \frac{\mu_o I}{4 \pi } [ \frac{\theta_{DA}}{r_{DA}} + \frac{\theta_{BC}}{r_{BC}}+ \frac{\theta_{CD}}{r_{CD}}+\frac{\theta_{AB}}{r_{AB}}]$

Where

$\theta_{BC} = \theta_{DA} = \frac{2\pi}{3}$

Where $\frac{2 \pi}{3} = 120^o$

$\theta_{CD} = \theta_{AB} = 0^o$

so

$B = \frac{\mu_o I}{4 \pi } [ \frac{\frac{2\pi}{3} }{0.20} - \frac{\frac{2\pi}{3}}{0.30}}+ \frac{0}{0.10}+\frac{0}{0.10}]$

$B = 3.87 *10^{-6} \ T$

The direction is into the page

This because the magnitude of the magnetic field due to arc BC whose direction is outward is less than that of DA whose direction is inward

This is because according to Fleming's Left Hand Rule the direction of current is perpendicular to the direction of magnetic field so since current in arc BC and DA are moving in opposite direction their magnetic field will also be moving in opposite direction