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Ber [7]
3 years ago
13

11y^2-19y-10=-4y^2 all real solutions

Mathematics
1 answer:
tangare [24]3 years ago
4 0

Answer:

y=5/3 is the only real solution

Step-by-step explanation:

Solve for y over the real numbers:

11 y^2 - 19 y - 10 = -4 y^2

Add 4 y^2 to both sides:

15 y^2 - 19 y - 10 = 0

The left hand side factors into a product with two terms:

(3 y - 5) (5 y + 2) = 0

Split into two equations:

3 y - 5 = 0 or 5 y + 2 = 0

Add 5 to both sides:

3 y = 5 or 5 y + 2 = 0

Divide both sides by 3:

y = 5/3 or 5 y + 2 = 0

Subtract 2 from both sides:

y = 5/3 or 5 y = -2

Divide both sides by 5:

Answer: |

| y = 5/3 or y = -2/5

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stira [4]

Firstly, add 2 to both sides of the equation: \frac{3}{5}x^2=-3

Next, multiply both sides by 5/3: x^2=-5

Next, square root both sides: x= \pm \sqrt{-5}

Lastly, factor out i (i = √-1) and <u>your answer will be x= \pm \sqrt{5}i</u>

7 0
3 years ago
Sandra​, who is paid​ time-and-a-half for hours worked in excess of 40​ hours, had gross weekly wages of $ 608 or 56 hours worke
vodomira [7]
Time-and-a-half doesn't make enough sense to answer the question. If you fix it, or comment below the correction I will gladly help.
4 0
3 years ago
"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e
Art [367]

Answer:

a. v(t)= -6.78e^{-16.33t} + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=e^{\int\limits^  {}k \, dt } =e^{kt}. We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ e^{kt}v' + ke^{kt}v = ge^{kt} ⇒ [ve^{kt}]' = ge^{kt}. Integrating, we have

∫ [ve^{kt}]' = ∫ge^{kt}

    ve^{kt} = \frac{g}{k}e^{kt} + c

    v(t)=   \frac{g}{k} + ce^{-kt}.

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + ce^{-16.33 * 0} = 16.33 + c

       c = 9.55 -16.33 = -6.78.

So, v(t)=   16.33 - 6.78e^{-16.33t}. m/s = - 6.78e^{-16.33t} + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78e^{-16.33 x 0.5} + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0
3 years ago
What is the answer to 2937682+28373938
xxMikexx [17]

Answer: 31311620

Step-by-step explanation: Calculator

5 0
2 years ago
Can someone help 12 points on the line
ANTONII [103]

Answer:

x=6

Step-by-step explanation:

We can write this as a ratio

x        9

---- = --------

10       15

Using cross products

15x = 10*9

15x = 90

Divide by 15

15x/15 = 90/15

x =6

7 0
3 years ago
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