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Sauron [17]
3 years ago
8

Find the odd one out:

Mathematics
2 answers:
lys-0071 [83]3 years ago
8 0
C. Tree: Forest, because the rest is the group, then what it is made up og, for example, a class is made up of students, but a tree is not made up of a forest
iris [78.8K]3 years ago
3 0
I believe it would be C. Tree: Forest.

A, B, and D is all whole to part, but C is part to whole.
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What measures of the cylinder do 26 and 34 describe?
dezoksy [38]

radius and diameter

The Radius is the distance from the center outwards. The Diameter goes straight across the circle, through the center. The Circumference is the distance once around the circle.

5 0
3 years ago
Read 2 more answers
I need some help on this question ASAP
telo118 [61]

Answer:

I would believe the answer is A

6 0
3 years ago
Someone pls help im new to this so tysm if you do help:D
tensa zangetsu [6.8K]

Answer:

-6

Step-by-step explanation:

8 0
3 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 = 2.0ms^-2 and  

g2 = 9.8ms^-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 = 2.0ms^-2  and g2 = 9.8ms^-2

We get h1 = 3.67m which is the required height

6 0
4 years ago
I neeeeeeeeeeed help
Musya8 [376]
He gave away 14 bouncy balls
3 0
3 years ago
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