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3 years ago

How do you work out the surface area of a cuboid?

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Surface area=
(width×length+width×height+height×length)×2

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a juggler throws a ball into the air, releasing it 5 feet above ground with an initial velocity if 15 fps. she catches the ball

Lena [83]

Problems of this sort are frequently found in physics. If you study calculus or physics you'll learn how to create the equation representing the velocity of an object in flight.

Here, you don't need to calculate velocity, but rather time. Start with this equation:

v = v0 + a t^2, where v is the velocity at time t, v0 is the initial velocity, a is the acceleration due to gravity (denoted by g instead of a), and t is the elapsed time.

You are told that v0 is 15 ft/sec. Set v = to 0, as the ball stops moving for the tiniest instant at the top of its trajectory. Use g = - 32 ft (per second squared).

Then 0 = 15 ft/sec - 32 [ft/(seconds squared)] t.

Solve this for t. This is the time required for the ball to come to a complete stop at the top of its trajectory.

Finally, multiply this time by 2, since the ball begins to fall and returns to its original height.

5 0

4 years ago

3/-4 divided by (-2 1/3)

Butoxors [25]

Answer:

Exact Form = $\frac{9}{28}$

Decimal Form = 0.32142857

3 0

3 years ago

Read 2 more answers

Please solve this I don’t understand it

GrogVix [38]

Answer:

y= 9

Step-by-step explanation:

-2 + 9 =7

start at -2 go to the postive side by 9 and you land on 7

idk # 2 tho sorry

4 0

4 years ago

Solve the equation after making an appropriate substitution. X^6-98x^3=3375

Leni [432]

Rearrange $x^6-98x^3=3375$ to give

$x^6-98x^3-3375=0$

Let $x^3$ be a variable 'p' and so we can write $x^6$ as $x^6=(x^3)(x^3)=(p)(p)= p^2$

Rewrite the equation in terms of 'p'

$p^2-98p-3375 = 0$

where $a=1, b=-98, c=-3375$

using the quadratic formula $\frac{-b+ \sqrt{b^2-4ac} }{2a}$ and subsitute the value of $a, b, c$

$p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125$

$p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27$

There are two value of p; 125 and -27

Now we find the value of x

Earlier we substitute $x^3$ for $p$ and $x^6$ for $p^2$

When $p=125$ , $x= \sqrt[3]{125}=5$
When $p = -27, x= \sqrt[3]{-27}=-3$

So the final answer is the two values of x;

x = 5 OR x = -3

3 0

3 years ago

PLEASE HELPPPPPP. HAS TO HAVE AN EXPLANATION.

Maurinko [17]

Answer:

Hello! Your answer would be BELOW!

Step-by-step explanation:

1: A)

2: D)

3: D)

4: C)

Hope I helped! Ask me anything if you have more questions! Have a nice mourning! Brainiest plz! Hope you make an 100%! -Amelia♥

4 0

3 years ago