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Pie
2 years ago
10

A formula for the normal systolic blood pressure for a man age a , measured in mmhg, is given as p=0. 006a2−0. 02a 120. Find the

age of a man whose normal blood pressure measures 123 mmhg
Physics
1 answer:
djyliett [7]2 years ago
5 0

The age of a man whose normal blood pressure measures 123 mm of hg

9 years

<h3>What is Quadratic equation ?</h3>

A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic equation is y = ax^{2} + bx + c, where a, b, and c are numbers and a cannot be 0

P(A) = 0.006 a^{2} - 0.02a + 120

123 = 0.006- 0.02a + 120

0=0.006 a^{2} - 0.02a - 3

you can use the quadratic equation  formula to solve for the man's age.

A = (-b ± (\sqrt{b^{2} - 4*a*c})  ) / (2a)

A = (0.02 ±  \sqrt{(-0.02)^{2} - 4*0.006*(-3)}/ (2*0.006)

A = (0.02 ± \sqrt{0.0076}) / 0.012

A = 9 , -5.67

Age of the man will be 9 years

To learn more about quadratic equation  here

brainly.com/question/17177510?referrer=searchResults

#SPJ4

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Watt-hours is a measure of energy, just like kilowatt-hours. How can you convert this to Joules?
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3 years ago
A magnetic field is directed perpendicular to the plane of a 0.15-m × 0.30-m rectangular coil consisting of 240 loops of wire. T
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7.344 s

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6 0
3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

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The charge chould be moved to infinity

7 0
3 years ago
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