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3 years ago

Ms Hogan has more than 20 students in her class.

Mathematics

1 answer:

ELEN [110]3 years ago

7 0

Answer:

The answer is most likely 29

Step-by-step explanation:

you can have 5 groups of 5 with 4 left over, and 7 groups of 4 with one left over.

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If 15% of a container of milk is white milk what is the fraction of the remaining chocolate milk

Fittoniya [83]

Answer:

17/20

Step-by-step explanation:

15/100= white milk

85/100= chocolate milk

85/5=17

100/5=20

=17/20 is chocolate milk

4 0

3 years ago

Deleteddddddddddddddddd

Sonbull [250]

Answer:

i do not think it is possible to delete a question after you have posted it. if i were you i probably would have just edited it and save myself the emmbarasment

Step-by-step explanation:

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2 years ago

PLEASE HELP

devlian [24]

Answer:

I think it is a 160o Angle

Step-by-step explanation:

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=5%20%2B%204%20-%20%28%205%20-%20%286%20-%205%29%29" id="TexFormula1" title="5 + 4 - ( 5 - (6 -

Ket [755]

(6-5)=1 (5-1)=4 5+4=9 9-4=5

8 0

3 years ago

Read 2 more answers

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

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

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

A) Since the maximum jump height is given by the formula

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

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) Formula to find height of ball thrown is given by

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

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