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tensa zangetsu [6.8K]
3 years ago
15

What is an expression for the distance between the origin and a point P(x,y)

Mathematics
1 answer:
kirza4 [7]3 years ago
5 0

Answer:

d =  \sqrt{ {x}^{2}  +  {y}^{2} }

Step-by-step explanation:

Distance between origin (0, 0) and point (x, y) is given as:

d =  \sqrt{ {(x - 0)}^{2}  +  {(y - 0)}^{2} }  \\  \\ \red{ \bold{ d =  \sqrt{ {x}^{2}  +  {y}^{2} } }}

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The diameter of a bacteria colony that doubles every hour is represented by the graph below. What is the diameter of the bacteri
Nady [450]

Do you notice anything strange about those points ?

(0, 1), 
(1, 2), 
(2, 4),
(3, 8).

The y-coordinate of each point is (2) raised to the  power of the x-coordinate.

First point:       x=0,  y=2⁰ = 1
Second point: x=1,  y=2¹ = 2
Third point:      x=2,  y=2² = 4
Fourth point:    x=3,  y=2³ = 8


The equation of the curve appears to be

                               y = 2 ^ x .

So, after 10 hours,  x=10, and  y = 2¹⁰ = 1,024 .

3 0
3 years ago
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Given g(x)=-2x-1g(x)=−2x−1, find g(-4)g(−4)
nydimaria [60]

Answer:

7

Step-by-step explanation:

4 0
3 years ago
A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
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What is a ratio called with two equivalent measurements?
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Ratio with two equivalent measurements would be 1
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