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3 years ago

A. Transform F using the rule (x, y) → (-x, y)

Mathematics

1 answer:

IceJOKER [234]3 years ago

3 0

Answer:

A: Use these points to plot the shapes reflection (-3,4) (-3,6) (-5,6) (-7,4)

B: The transformation would be described as a reflection over the y axis

C: The transformation does result in a congruent figure because the shape doesn't change in size or shape and in length or width

Step-by-step explanation:

So our plots from the figure are: (3,4) (3,6) (5,6) (7,4)

This rule (x, y) → (-x, y) is used for the type of transformation that is a reflection but over the y axis.

So using the rule (x, y) → (-x, y) are new points would be:

(-3,4)

(-3,6)

(-5,6)

(-7,4)

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Plsssss help I'm begging plsss i have a math exam

IceJOKER [234]

Answer:

what is the question you are talking about ?

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3 years ago

Find the seventh term of the

jekas [21]

Answer:

$T_7 = \frac{64}{729}$

Step-by-step explanation:

Given

$a =1$

$r = \frac{2}{3}$

Required

Determine the 7th term

The nth term of a gp is:

$T_n = a * r^{n-1$

So, we have:

$T_7 = 1 * \frac{2}{3}^{7-1$

$T_7 = 1 * \frac{2}{3}^{6$

$T_7 = 1 * \frac{2^6}{3^6}$

$T_7 = 1 * \frac{64}{729}$

$T_7 = \frac{64}{729}$

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3 years ago

I need the domain to this.

aliina [53]

Answer:

all real numbers greater than zero

Step-by-step explanation:

The square root of a number must be greater than or equal to zero

sqrt(x) ≥0

Squaring each side

x ≥0

This means the domain is all real numbers greater than or equal to zero

I would go with all real numbers greater than zero since it is the closest answer

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3 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

4 years ago

This is 20 points if you don't know the answer please don't put your question down thank you...

MissTica

Answer:

I believe it is C) They both involve writing a rate.

hope this helps

Step-by-step explanation:

6 0

3 years ago

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