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Alja [10]
2 years ago
8

If O is an angle in standard position and its terminal side passes through the point

Mathematics
1 answer:
Nataliya [291]2 years ago
6 0

Answer:

-1/3

Step-by-step explanation:

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How do u subtract (3x+5y-4)_(4×+11)
AleksAgata [21]
Distribute the negative sign to the 4x and 11 making it 3x+5y-4-4x-11 and combine like terms 
5 0
3 years ago
Write the equation of a possible rational
kondor19780726 [428]

Answer:

The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)

Step-by-step explanation:

A rational equation is a equation where

\frac{p(x)}{q(x)}

where both are polynomials and q(x) can't equal zero.

1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of

(x - 2)

in our denomiator.

So right now we have

\frac{p(x)}{(x - 2)}

2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.

We can model -2.5 as

(2x + 5)

So we have as of right now.

\frac{(2x + 5)}{(x - 2)(2x + 5)}

Now let see if this passes throught point (6,-3).

\frac{(2x + 5)}{(x - 2)(2x + 5)}  = y

\frac{(17)}{68}  =  \frac{1}{4}

So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.

If we have a variable r, in the numerator that will make this applicable, we would get

\frac{(2x + 5)r}{(2x + 5)(x - 2)}  =  - 3

Plug in 6 for the x values.

\frac{17r}{4(17)}  =  - 3

\frac{r}{4}  =  - 3

r =  - 12

So our rational equation will be

\frac{ - 12(2x + 5)}{(2x + 5)(x - 2)}

or

\frac{ - 24x - 60}{2 {x}^{2}  + x - 10}

We can prove this by graphing

5 0
2 years ago
Determine the remaining sides and angles of the triangle ABC
Lapatulllka [165]

Answer:

The measure of angle B is 41.5°

The length of side a is 30.6 feet ⇒ to the nearest tenth

The length of side b is 32.9 feet ⇒ to the nearest tenth

Step-by-step explanation:

  • <em>The sum of the measures of the interior angles in a triangle is 180°</em>

In the given Δ ABC

∵ m∠A = 38.1°

∵ m∠C = 100.4°

→ By using the fact above

∵ m∠A + m∠B + m∠C = 180°

∴ 38.1 + m∠B + 100.4 = 180

→ Add the like terms in the left side

∴ 138.5 + m∠B = 180

→ Subtract 138.5 from both sides

∴ m∠B = 41.5°

∴ The measure of angle B is 41.5°

∵ a is the opposite side of ∠A

∵ b is the opposite side of ∠B

∵ c is the opposite side of ∠C

→ By using the sine rule

∴ \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

∵ c = 48.4 ft, m∠A = 38.1°, m∠B = 41.5°, and m∠C = 100.4

→ Substitute them in the sine rule above

∴ \frac{a}{sin38.1} = \frac{b}{sin41.5} = \frac{48.4}{sin100.4}

→ By using cross multiplication between the 1st and the 3rd fractions

∴ a × sin 100.4 = 48.8 × sin 38.1

→ Divide both sides by sin 100.4

∴ a = 30.61429861

∴ The length of side a is 30.6 feet ⇒ to the nearest tenth

→ By using cross multiplication between the 2nd and the 3rd fractions

∴ b × sin 100.4 = 48.8 × sin 41.5

→ Divide both sides by sin 100.4

∴ b = 32.87596206

∴ The length of side b is 32.9 feet ⇒ to the nearest tenth

4 0
3 years ago
A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi
sergij07 [2.7K]

Answer:

(a) After time (t), the amount of salt in the tank is s(t) = 130 − 130e ∧−3t/200 / 3

(b) After 60 minutes, the salt in the tank is s(60) ≈ 25.7153

Step-by-step explanation:

To start with,

Let s(t) = amount of salt in kg of salt at time t.

Then we have:

dt/ds =  (rate of salt into tank) − (rate of salt out of tank)

= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) − ( s kg/ 1000 L x 15 L/min)

= 0.25 kg/min + 0.4 kg/min − 15s / 1000 kg/min

So we get the differential equation

dt/ds = 0.65 − 15s / 1000

= 65 / 100 − 15s / 1000

dt/ds = 130 - 3s / 200

We separate s and t to get

1 / 130 - 3s ds = 1 / 200 dt

Then we Integrate,

Thus we have

∫1 / 130 - 3s ds = ∫1 / 200 dt

- 1/3  · ln |130 − 3s| =1 / 200 t + C1

ln |130 − 3s| = - 3 /200 t + C2

|130 − 3s| = e− ³⁺²⁰⁰ ∧ t+C2

|130 − 3s| = C3e − ∧3t/200

130 − 3s = C4e  ∧−3t/200

−3s = −130 + C4e  ∧−3t/200

s = 130 - C4e ∧−3t/200 / 3

Since we begin with pure water, we have s(0) = 0. Substituting,

0 = 130 − C4e  ∧−3·0/200 / 3

0 = 130 − C4

C4 = 130

So our function is

s(t) = 130 − 130e ∧−3t/200 / 3

After one hour (60 min), we have

s(60) = 130 − 130e ∧−3·60/200 / 3

s(60) ≈ 25.7153

Thus, after one hour there is about 25.72 kg of salt in the tank.

6 0
2 years ago
If 3x - 7 = 5 then what is 6x -8
victus00 [196]
3x - 7 = 5
    + 7 + 7
     3x = 12
      3      3
       x = 4

6x - 8
6(4) - 8
24 - 8
16
7 0
3 years ago
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