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3 years ago

How do you estimate 12.062g to the nearest tens of a gram

Mathematics

1 answer:

gtnhenbr [62]3 years ago

6 0

12.602 = 10g
or if you meant tenTHs it will be 13g

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The shirt cost $113 less than the coat. The shirt cost $50. How much is the coat?

olga nikolaevna [1]

Shirt: $50
Coat: $113

$50+ $113= $163

The coat costs $163, hope this helps!

7 0

3 years ago

Explain why if a runner completes a 6.2â-mi race in 32 âmin, then he must have been running at exactly 11 âmi/hr at least twice

MA_775_DIABLO [31]

Answer:

Accelerating to top speed, deaccelerating to finish line.

Step-by-step explanation:

If the runner kept a constant speed of 11 mph for the whole duration of his run (32 minutes), the distance he would have covered is:

$d=11*\frac{32}{60}\\d= 5.87\ mi$

This means that, in order to run the full 6.2 miles, the runner needs to reach a speed over 11 mph. Assume he starts from rest, while accelerating the runner reaches, and the surpasses, the 11 mph mark. Since his speed at the finish line is zero, the runner has to deaccelerate from his current running speed (which should be higher than 11 mph), passing through 11 mph and reaching zero at the finish line.

3 0

3 years ago

A figure is shown with the given dimensions.

Illusion [34]

Answer:

$\blue{A = 24~ft^2}$

Step-by-step explanation:

This figure is a trapezoid with bases of 10 ft and 6 ft and height of 3 ft.

$A = \dfrac{b_1 + b_2}{2}h$

$A = \dfrac{10~ft + 6~ft}{2} \times 3~ft$

$\blue{A = 24~ft^2}$

4 0

3 years ago

What happens to light when it strikes a smooth shiny surface?

ELEN [110]

Step-by-step explanation:

Light reflects from a smooth surface at the same angle as it hits the surface. For a smooth surface, reflected light rays travel in the same direction. This is called specular reflection. For a rough surface, reflected light rays scatter in all directions.

6 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

Elena L [17]

Answer:

First option is the right choice.

Step-by-step explanation:

$\frac{1}{\cos \left(x\right)+1}+\frac{1}{\cos \left(x\right)-1}\\\\\frac{\cos \left(x\right)-1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}+\frac{\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{\cos \left(x\right)-1+\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{2cos\left(x\right)}{cos^2\left(x\right)-1}\\\\=\frac{2cos\left(x\right)}{sin^2\left(x\right)}$

$=2\cdot \frac{cos\left(x\right)}{sin\left(x\right)}\cdot \frac{1}{sin\left(x\right)}\\\\=-2cot(x)csc(x)$

Best Regards!

7 0

3 years ago