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kenny6666 [7]
3 years ago
13

The duration of a flash from a camera is timed at 0.004 seconds accurate to 1 / 1000 of a second calculate the upper and lower b

ounds for the duration of the flash using in quality
​
Mathematics
1 answer:
Agata [3.3K]3 years ago
7 0

Answer:

0.0035 ≤ x ≤ 0.0045

Step-by-step explanation:

Given that:

The error = 1/1000 = 0.001 / 2 = 0.0005 s

Hence, 0.004 ± Error

0.004 ± 0.0005

Lower boundary = (0.004 - 0.001) = 0.0035

Upper boundary = (0.004 + 0.001) = 0.0045

0.0035 ≤ x ≤ 0.0045

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Use the information provided to write the equation of each circle​
Nadusha1986 [10]

Answer:

1.

centre(h,k)=(-13,9)

radius (r)=6

we have

equation of the circle is

(x-h)²+(y-k)²=r²

(x+13)²+(y-9)²=6²

x²+26x+169+y²-18y+81=36

x²+y²+26x-18y+169+81-36=0

x²+y²+26x-18y +214=0

is a required equation of the circle.

2.

centre(h,k)=(1,-1)

radius (r)=11

we have

equation of the circle is

(x-h)²+(y-k)²=r²

(x-1)²+(y+1)²=11²

x²-2x+1+y²+2y+1=121

x²+y²-2x+2y=121-2

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is a required equation of the circle.

4 0
3 years ago
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



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