All categories

3 years ago

Can you please help me

Mathematics

1 answer:

skad [1K]3 years ago

4 0

Answer:

explanations:

Hope it helps!

You might be interested in

The graph of a function g is shown please find the following g(10)

Rama09 [41]

Check the picture below.

3 0

2 years ago

How to find the line of reflection on a graph?

Blababa [14]

Answer:

can you show us the graph

6 0

2 years ago

Determine whether each relationship is a function

Mademuasel [1]

It’s a function
hope you have a good day

8 0

2 years ago

Simply the following 2m⁵×5m³×7m²

Ganezh [65]

Answer:

$2m^5\cdot 5m^3\cdot 7m^2=70\cdot m^{10}$

Step-by-step explanation:

Simplify the expression

$2m^5\cdot 5m^3\cdot 7m^2$

Calculate the product of the coefficients:

2*5*7=70

Compute the product of the variables. Since they are all the same letter, sum their exponents:

$m^5\cdot m^3\cdot m^2=m^{5+3+2}=m^{10}$

Combine the results:

$\boxed{2m^5\cdot 5m^3\cdot 7m^2=70\cdot m^{10}}$

3 0

3 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

3 years ago