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16 days ago

Three friends went running. The table below shows their last 3 lap times.

Mathematics

1 answer:

tester [3.9K]16 days ago

5 0

Answer: 1 min 50 sec

Step-by-step explanation:

The difference is the maximum value minus the minimum value.

Minimum value = 14 mins 40 sec

Maximum value = 16 mins 30 sec

Thus, subtracting gives us 16 mins 30 sec - 14 mins 40 sec

= 1 min 50 sec

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Mrs. Adams bought cans of pears and cans of mixed fruit.

Svet_ta [4321]

Answer: 5 Cans

Step-by-step explanation:

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11 days ago

How many calories will Darrel burn in 1 minute while kayaking enter the answer below in the box 50/200, 100/400, 150/600, 200/80

Zina [3914]

Response: 100/400

Detailed explanation:

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11 days ago

Which shows the image of ΔRST after the rotation (x, y) → (y, –x)? On a coordinate plane, a triangle has points R (negative 2, 1

PIT_PIT [3919]

Answer:

On a coordinate grid, a triangle is defined by the points R' (1, 2), S' (3, -1), T' (7, 1)

Step-by-step explanation:

We can interpret the coordinates of the triangle's vertices as...

R(-2, 1), S(1, 3), T(-1, 7)

Applying the transformation (x, y) ⇒ (y, -x), these coordinates change to...

R'(1, 2), S'(3, -1), T'(7, 1) . . . . . (this corresponds to the first option)

4 0

16 days ago

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Erika is writing a coordinate proof to show that the diagonals of a rectangle are congruent. She begins by assigning coordinates

Leona [4166]

Answer:

Option C is the right choice.

Step-by-step explanation:

The given coordinates define a rectangle, and our objective is to show that the diagonals JL and KM are congruent.

We know that rectangles possess four right angles.

To prove the congruence of the diagonals JL and KM, we will utilize the Pythagorean theorem.

In triangle KLM, KL has a length of b units while LM has a length of a units. By applying the Pythagorean theorem $\sqrt(a^{2}+b^{2})=KM$

In triangle JML, JM is b units long, and LM remains a units long. We again can apply the Pythagorean theorem $\sqrt(a^{2}+b^{2})=JL$

Thus, we find that $\sqrt(a^{2}+b^{2})=KM=LM$ and option C is the correct choice.

7 0

4 days ago

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The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [3877]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$ g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

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11 days ago

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