You jog 3 kilometers in 12 minutes.

Mathematics

2 answers:

IgorLugansk [536]2 years ago

6 0

12/3 = 4 minutes
4 minutes per kilometre
4x7=28
28 minutes

Lana71 [14]2 years ago

5 0

Answer:

It will take 28 minutes to complete a 7-kilometer race

Step-by-step explanation:

If you can job 3 kilometers in 12 minutes, that means you can jog 1 kilometer in 4 minutes. so if you multiply 7 times 4, you would get 28. So your answer is 28 minutes.

You might be interested in

PLEASE HELP whats the hypotenuse of a right triangle if both its legs are 9 cm round to the nearest tenth of a cm

vivado [14]

Pythagorean theorem: a² + b² = c²

a² and b² are the legs, and c² is the hypotenuse.

Plug in 9 to a² and b² since you know that both the right triangle's legs are 9 cm.

9² + 9² = c²
81 + 81 = c²
162 = c²

Square root both sides.

c = 12.7 cm

6 0

3 years ago

Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) Compute Δy and dy for the

Darina [25.2K]

Base on the data you have given and with my calculation, i think the best value for Delta X and Dy is delta x = 2 and the DY = 1/2.I hope you are satisfied with my answer and feel free to ask for more if you have more clarifications and further questions

3 0

2 years ago

Read 2 more answers

In simplifying the expression 2( 3x + 5) -6, what is the first thing to do?

hodyreva [135]

Answer:

distribute 2 inside the parenthesis

Step-by-step explanation:

3 0

2 years ago

Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$