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3 years ago

Translate the sentence into an equation.

Mathematics

1 answer:

maks197457 [2]3 years ago

4 0

Answer:

9y+8=5

Step-by-step explanation:

1. "Times" means multiply and "and" means add

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Hey, can yall help me with this question?

RSB [31]

SIR YOUR ANSWER IS B. I K FO SHO. THANK YOU. HOPE THIS HELPS. MIND MARKING BRAINLIEST AGAIN? XD THANKS

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3 years ago

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Dominik [7]

Answer:

Step-by-step explanation:

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3 years ago

What is the slope of a line that passes through (2, 77) and (98, 5)

blagie [28]

Answer:

Step-by-step explanation:

slope = (5-77)/(98-2) = -72/96 = -¾

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3 years ago

marys age is 2 upon 3 that of peters age. two years ago marys age was 1 upon 2 of whats peters age will be in 5years time. how o

Nitella [24]

Answer:

the Peter age be 27

Step-by-step explanation:

Let assume peter be

And, then mary is 2p ÷ 3

The equation would be

2p ÷ 3 - 2 = 1 ÷ 2 (p + 5)

4p - 12 = 3p + 15

p = 27

So Peter age be 27

And, the mary age is

= 2 (27) ÷ 3

= 18

Hence, the Peter age be 27

4 0

2 years ago

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