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3 years ago

What is the surface area of the right square pyramid shown above ?

Mathematics

2 answers:

Sergio [31]3 years ago

8 0

Answer:

your answer will most likely be D)

Art [367]3 years ago

5 0

Answer:

answer d is the right answer

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Somebody please help with this math problem.19=4b-13

Veronika [31]

Start by adding 13 to both sides

19 = 4b-13
19+13 = 4b-13+13
32 = 4b
4b = 32

Then divide both sides by 4 to isolate b

4b = 32
4b/4 = 32/4
b = 8

So in the end, b = 8

5 0

4 years ago

Read 2 more answers

To do his homework for a drafting class, Axel needs a round table with an area of at least 45ft. What is the area of his parent’

Natasha2012 [34]

Answer:

About 38.48ft

Step-by-step explanation:

A= $\pi$ $r^{2}$

A= $3.5^{2}$ $\pi$

A≈38.48

8 0

3 years ago

One angle is 36 Degrees. What is the measure of its complement?

Andre45 [30]

Complementary angles equal 90°

So 90-36= 54°

4 0

3 years ago

Help Find the volume of this sphere. Use 3 for pi

tankabanditka [31]

Answer:

V = 1372cm3

Step-by-step explanation:

See the image above

7 0

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