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3 years ago

HURRY PLS i’ll give brainliest please help thanks :)

Mathematics

1 answer:

lianna [129]3 years ago

5 0

Answer:

67°F

Step-by-step explanation:

The mean is another way to say average.

To find the average: you have to add up all of the temperatures, then divide by how many numbers (in this case, time slots) are there.

58°F + 60°F + 61°F + 67°F + 69°F + 73°F + 74°F + 74°F = 536°F

536°F ÷ 8 (total number of time slots) = 67°F

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s2008m [1.1K]

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Step-by-step explanation: 48f does not equal 8.6f

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3 years ago

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Korolek [52]

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3 years ago

Please help <3 Paula is writing short, casual notes for how to find the inverse of a function, f. Those notes are shown below

NARA [144]

Answer:

(3rd listed choice) Paula's notes are incorrect. Step 4a should be "If the result does not define y as a function of x, then f does not have an inverse," and step 4b should be "If the result defines y as a function of x, then f has an inverse."

Step-by-step explanation:

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6 0

3 years ago

(20 points)

Nuetrik [128]

Answer:

the answer is B. 99x + 84 ≥ 1,150

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago