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3 years ago

Find the exact solutions of the equation x2 +3x + 1=0.

Mathematics

2 answers:

adell [148]3 years ago

7 0

X =(-3-√13)/2=-3.303, or <span>x =(-3+√13)/2= 0.303 either one of these should work, hope this helps :)</span>

zimovet [89]3 years ago

4 0

X2+3x+1=0

5x+1=0
-1 -1
5x=-1
/5 /5
x=-0.2

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A pilot has just started on the glide path for landing at an airport with a runway of length 9000 feet. The angles of depression

ki77a [65]

Answer:

a. See attachments

B. 163630.2ft

C. 147056.70ft

D. 49205.4ft

Step-by-step explanation:

Check attachments for details

Angle a.. Angle of dep to beginning of runway

Angle b... Angle of del to end of runway

AD... Altitude of plane

DB..... Ground distance before touch down

AC... Air distance travelled until touch down on the near end of runway

5 0

3 years ago

(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi

Drupady [299]

Answer:

a. $\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. $x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. $c = \dfrac{3}{8} \ g/L$

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx = (5·c - 5·c/3)×dt = 20/3·c = $6\frac{2}{3} \cdot c \cdot dt$

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. The amount of salt, x after t minutes is given by the relation

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

$dx = 6\frac{2}{3} \cdot c \cdot dt$

$x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt$

$x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

$x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10$

$6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}$

$c =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} } = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200} \ g/L = \dfrac{3}{8} \ g/L$

$c = \dfrac{3}{8} \ g/L$

4 0

3 years ago

1. Use the Law of Levers to solve the problem, x1y1=x2y2.

Mademuasel [1]

1) Never heard of someone referring to the Law of Levers before, must be showing my age. A system as you describe is at equilibrium when the net torque is equal to zero. Since torque is Fr=mgr so for this system we have a sum of:

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270+30x=0

270=-30x

x=-9

The negative just indicates that the magnitude of 9 is on the opposite side of the fulcrum, so x=9ft in magnitude.

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y=k/x, and we are given y=-12 when x=5 so

-12=k/5

k=-60 so the equation is:

y=-60/x which can be expressed as

xy=-60

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3 years ago

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Simora [160]

This one looks tricky let me find out

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3 years ago

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Blababa [14]

-2: y = -2-2 = -4
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0: y = 0 - 2 = -2
1: y = 1 - 2 = -1
2: y = 2 - 2 = 0

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2 years ago