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Anna11 [10]
2 years ago
7

Find all of the points of the form (x, −x) which are 7 units from the origin.

Mathematics
2 answers:
aksik [14]2 years ago
6 0

The equation will be

\\ \sf\longmapsto \sqrt{x^2+y^2}=7

\\ \sf\longmapsto x^2+y^2=7

  • If y be 0

\\ \sf\longmapsto x^2+0=7^2

\\ \sf\longmapsto x^2=49

\\ \sf\longmapsto x=\sqrt{49}

\\ \sf\longmapsto x=\pm 7

Now the points are

  • (7,0)
  • (-7,0)
Morgarella [4.7K]2 years ago
3 0

Answer:

  • (-7, 0) and (7, 0)

Step-by-step explanation:

<u>All the points that are 7 units from the origin are on the circle:</u>

  • x² + y² = 7²

<u>The maximum and minimum values of x obtained when y = 0:</u>

  • x² = 7²
  • √x² = √7²
  • x = ± 7

So

  • min x = -7 and
  • max x = 7

<u>The points are:</u>

  • (-7, 0) and (7, 0)
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Simplify the following expression. 3 2/5x3(-7/5)
Studentka2010 [4]

The simplified expression of 32/5 x 3(-7/5) is  -672/25

<h3>How to simplify the expression?</h3>

The mathematical expression is given as:

32/5 x 3(-7/5)

Evaluate the product of 3 and -7/5

32/5 x 3(-7/5) =  32/5 x -21/5

Evaluate the product of 32 and -21

32/5 x 3(-7/5) =  672/5 x -1/5

Evaluate the product of 5 and 5

32/5 x 3(-7/5) =  672/25 x -1

So, we have:

32/5 x 3(-7/5) =  -672/25

Hence, the simplified expression of 32/5 x 3(-7/5) is  -672/25

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-2 2/3, -5 1/3, -10 2/3, -21 1/3, -42 2/3, ... Which formula can be use to describe the sequence?
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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
3 years ago
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