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4 years ago

What is the distance between the points (3, 7) and (15, 16) on a coordinate plane?

Mathematics

1 answer:

EastWind [94]4 years ago

5 0

Answer:

4.123. Rounded to the nearest tenth, you get 4.1 units.

Step-by-step explanation:

I've included a snapshot of my work.

I hope this helps. If you have any more questions, please feel free to post them and someone will be able to help you, whether it's myself or others. Please leave a like, rating, and if possible, Brainliest. Have a great day!

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a baker cut a pie in half. he cut each half into 3 equal pieces and each piece into 2 equal slices. he sold 6 slices. what fract

olga nikolaevna [1]

Answer:

1/2 is sold

Step-by-step explanation:

7 0

3 years ago

NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t2+121t

Alenkasestr [34]

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by

$h(t)=-4.9t^2+121t+283$

The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:

$h(t)=0\implies-4.9t^2+121t+283=0$

Using the quadratic formula, the solution for our problem is

$\begin{gathered} t=\frac{-(121)\pm\sqrt{(121)^2-4(-4.9)(283)}}{2(-4.9)} \\ =\frac{121\pm142.083778...}{9.8} \\ =\frac{121+142.083778...}{9.8} \\ =26.8452834694... \end{gathered}$

The rocket splashes after 26.845 seconds.

The maximum of this function happens at the root of the derivative. Differentiating our function, we have

$h^{\prime}(t)=-9.8t+121$

The root is

$-9.8t+121=0\implies t=\frac{121}{9.8}=12.347$

Then, the maximum height is

$h(12.347)=1029.99$

1029.99 meters above sea level.

6 0

1 year ago

A thermometer is taken from a room where the temperature is 20◦C to the

lord [1]

Answer:

a) $T(2) = 8.265$ : at time t = 2 minutes the temperature will be 8.265 degress

b) $6 = T(3.55)$ : the temperature will be 6 degrees at time t = 3.55 minutes.

Step-by-step explanation:

When dealing with temperature changes, it's best to work with Newton's Law of Cooling.

$T(t) = T_s + Ce^{kt}$

here:

$T(t)$ : the temperature in the room.

$T_s$ : ambient (or outdoor) temperature (that always remains constant, in our case: $T_s = 5$ )

$C\,\text{and}\,k$ : are constants

Our conditions are provided:

1) $T(0) = 20$

2) $T(1) = 12$

using the first condition

$T(0) = 5 + Ce^{k(0)}\\20 = 5 + C(1)\\C = 15$

using the second condition:

$T(1) = 5 + Ce^{k(1)}\\12 = 5 + Ce^{k}\\e^k = \dfrac{7}{C}$

we can use our calculated value of C to find k

$e^k = \dfrac{7}{15}\\k = \ln{(\dfrac{7}{15})}\\k = -0.7621$

Finally we can put these constants back in the main equation:

$T(t) = T_s + Ce^{kt}$

$T(t) = 5 + 15e^{-0.7621t}$ or $T(t) = 5 + 15e^{\ln{(\frac{7}{15})t}$

a) Reading after one more minute?

so it's asking:

T(2) = ?

$T(2) = 5 + 15e^{\ln{(\frac{7}{15})(2)}}\\T(2) = \dfrac{124}{15} \approx 8.267$

Hence, after one more minute the temperature of the room will be 8.267 degrees

b) When will it be 6 degrees?

T(t) = 6?

$6 = 5 + 15e^{-0.7621t}\\\text{and solve for $t$}\\\\\dfrac{6-5}{15}=e^{\ln{(\frac{7}{15})}t}\\\ln{\left(\dfrac{1}{15}\right)} = \ln{\left(\dfrac{7}{15}\right)t}\\\ln{\left(\dfrac{1}{15}\right)} \div \ln{\left(\dfrac{7}{15}\right)} = t \approx 3.55\\$