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3 years ago

14

a cube has edge length 10 cm. starting at the vertices dots are placed along every edge at 1cm intervals. how many dots will the

re be in total once the pattern has been completed?

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

120 dots

Step by step explanation:

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A value of a car is $24,000. You pay it off by making payments of $6200 a year. Write and equation for when the car would be pay

Damm [24]

Answer: 24,000 / 6,200 = x years

Step-by-step explanation:

The car is valued at $24,000.

Yearly payments are $6,200.

The year that it will be paid off can be dtermined by calculating how long it would take $6,200 to get to $24,000.

You can do that by dividing $24,000 by $6,200.

Assume the number of years for the car to be paid off is x:

24,000 / 6,200 = x years

<em>When solved:</em>

<em>= 24,000 / 6,200</em>

<em>= 3.87 years </em>

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2 years ago

Which expression is equivalent to this expression?

saw5 [17]

Answer:

3h-4.5

Step-by-step explanation:

3/4(4h-6)

12/4h-18/4

3h-4.5

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4 0

2 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

<u>Answer:</u>

a) 3.675 m

b) 3.67m

<u>Explanation:</u>

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula </u>

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by </u>

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

3 years ago

I REALLY NEED HELP PLEASE 30 POINTS AND BRIANLIEST

Mademuasel [1]

Answer:

C is the closest one but it should be -472 instead -475

Step-by-step explanation:

just insert X in the formulas and compare to P

7 0

3 years ago

Find the missing lengths:<br> LO=5 and OK=4, find OH and KH.

hodyreva [135]

Answer:

The length of KH is 6 units and OH is 6.3 units.

Step-by-step explanation:

Given the figure with lengths LO=5 and OK=4. we have to find the length of OH and KH.

In ΔLOH

By Pythagoras theorem

$LH^2=LO^2+OH^2\\\\LH^2=5^2+OH^2$ → (1)

In ΔKOH,

$KH^2=OH^2+OK^2\\\\KH^2=OH^2+4^2$ → (2)

In ΔKHL,

$KL^2=LH^2+KH^2$

Using eq (1) and (2), we get

$KL^2=5^2+OH^2+OH^2+4^2$

$9^2=25+2OH^2+16$

⇒ $2OH^2=81-25-16=40$

⇒ $OH=\sqrt{20}=6.324\sim6.3units.$

Put the above value in eq 2, we get

$KH^2=20+4^2=36$

⇒ KH=6 units.

8 0

3 years ago

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