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Umnica [9.8K]
2 years ago
5

Translated up 3 units and left 2 units

Mathematics
1 answer:
seraphim [82]2 years ago
6 0
What are we trying to answer here lol
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Solve the simultaneous equation y-2x+1=0 and 4x^2+3y^2-2xy=7
Lapatulllka [165]

The first equation is y-2x+1=0

y=2x-1 (Equation 1)

The second equation is 4x^{2}+3y^{2}-2xy=7 (Equation 2)

Putting the value of x from equation 1 in equation 2.

we get,

4x^{2}+3(2x-1)^{2}-2x(2x-1)=7

4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7

by simplifying the given equation,

12x^{2}-10x-4=0

6x^{2}-5x-2=0

Using discriminant formula,

D=b^{2}-4ac

D=25-4 \times 6 \times -2 = 73

Now the formula for solution 'x' of quadratic equation is given by:

x=\frac{-b+\sqrt{D}}{2a}  and  x=\frac{-b-\sqrt{D}}{2a}

x=\frac{5+\sqrt{73}}{12}  and x=\frac{5-\sqrt{73}}{12}

Hence, these are the required solutions.

8 0
3 years ago
Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0
3 years ago
From a point 80 feet from the base of a
VMariaS [17]

Answer:10151 will be your answer

Step-by-step explanation:

6 0
3 years ago
Find the median of the following set of data. 13, 11, 4, 5, 6, 9, 10, 12, 15, 16
klemol [59]
Firstly arrange data from smaller to greater data point:
4 5 6 9 10 11 12 13 15 16
Secondly find the middle value: In our case the number of data is even, so to get the middle value we have to find the mean (average) between 10 & 11, that is 10.5. So the rank of the Median is10.5
8 0
3 years ago
What is 45 as a fraction?
Rama09 [41]
A whole number is always written as itself over one.

45/1 

Answer:   45/1
7 0
3 years ago
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