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2 years ago

Trapezoid JKLM is shown on the coordinate plane below

Mathematics

2 answers:

Anna11 [10]2 years ago

8 0

Answer:

b i think (im not really sure

Step-by-step explanation:

photoshop1234 [79]2 years ago

4 0

L (3, -2)
L’ (3 + 5, -2 - 4)
L’ (8, -6)
(C)

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0.66x10 to the 3rd power

Bas_tet [7]

The answer to this is 660

4 0

2 years ago

Read 2 more answers

A trapezoid is 64cm^2. The height is 12 cm and one base is 4cm. What is the length of the missing base?

Debora [2.8K]

Answer:

20/3

Step-by-step

(base1+base2)/2 * height = area of a trapezoid

64/12=16/3

16/3*2 = 32/3

32/3-4 = 20/3

4 0

3 years ago

Please help :( im struggling

Paha777 [63]

Answer:

a= 9/4

Step-by-step explanation:

2m^(5/2)/m^(1/4)

2m^(5/2 -1/4)

2m^(9/4)

Hope you find this helpful <3

8 0

2 years ago

How do I do this please help me I will give you thank even tho it’s not important Ahahah please help me

xxTIMURxx [149]

Answer:

(-2 , 5)

(-1 , 0)

(1 , -4)

(3 , 0)

(4 , -5)

Step-by-step explanation:

First solve the equation:

x² - 2x - 3

Find two numbers with have a sum of -2 and a product of -3.

-3 and 1

(x - 3)(x + 1)

Solve for x:

x - 3 = 0

x = 3

x + 1 = 0

x = -1

You know that the graph will cross the x-axis at -1 and 3.

(-1 , 0)

(3 , 0)

You know that the graph is positive.

Complete the square to find the vertex

x² - 2x - 3

(x - 1)² = x² - 2 + 1

x² - 2x - 3 = x² - 2 + 1 - 2 = (x - 1)² - 2

1 = 0

x = 1

Substitute into the original equation:

x² - 2x - 3 =

1² - (2 * 1) - 3 =

1 - 2 - 3 =

-4

(1 , -4)

You can input any two numbers within -10 and 10. Such as -2 and 4.

x² - 2x - 3 =

-2² - (2 * -2) - 3 =

4- -4- 3 =

(-2 , 5)

x² - 2x - 3 =

4² - (2 * 4) - 3 =

16 - 8 - 3 =

-5

(4 , -5)

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