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dlinn [17]
3 years ago
7

Noah and Diego left the amusement park’s ticket booth at the same time. Each moved at a constant speed toward his favorite ride.

After 8 seconds, Noah was 17 meters from the ticket booth, and Diego was 43 meters away from the ticket booth.
WILL GIVE BRAINEST IF YOU TELL ME how can you tell Diego’s is the blue line and Noah’s is the black?

Mathematics
1 answer:
lana [24]3 years ago
6 0

Answer:

because when Noah was 17 Diego was 43 so his is higher so his would be more upstraight than Noah's

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The side of one square is equal to 3m and its diagonal is equal to the side of a second square. Find the diagonal of the second
navik [9.2K]

Answer:

6m

Step-by-step explanation:

the side of first square = 3 m

area of first square = 3^2= 9 m^2

let the side of second square = x m

since diagonal of first square = x m as per the question

in a square each angle is 90 degree

therefore by applying pythagorus theorem ,

diagonal ^2 = side^2 + side ^2

diagonal^2 = 3^2+3^2=9+9

DIAGONAL= square root of 18

diagonal= 3\sqrt{2} m = side of second square

therefore in second square ,

one angle is 90  degree , therefore by applying pythagorue theorem

diaqonal^2 = side ^2 + side ^2 = (3\sqrt{2})^2 + ( 3\sqrt{2})^2

diagonal ^2 = 18 + 18

diagonal = square root of 36

therefore diagonal of second square is 6m

hope it helps you... please mark me as the brainliest..

8 0
3 years ago
Read 2 more answers
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

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3 years ago
In the base-ten number system, ten digits {0, 1, 2, 3, 4, 5, 7, 8, 9) are used for number
SVETLANKA909090 [29]

Answer:

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