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Otrada [13]
3 years ago
8

The second one write it on the paper and take a picture

Mathematics
2 answers:
Elodia [21]3 years ago
6 0

Answer:

The common for the second question is 30

Sati [7]3 years ago
6 0

Answer:

39/30 Here 30 will be common in denominators

Step-by-step explanation:

6/15×2/2=12/30

3/10×3/3=9/30

3/5×6/6=18/30

= 12/30+9/30+18/30

39/30

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How much material is needed to construct a triangular tent that is 6 feet wide 4 feet tall and 8 feet long with side measurement
jonny [76]

Answer: 152ft^2

The amount of material needed to construct the tent is 152ft^2

Step-by-step explanation:

The amount of material needed to construct a triangular tent= the surface area of the triangular tent.

S.A = area of the 2 side rectangles + area of the 2 side triangles + area of the base.

S.A = Ab + At + Ar ....1

Area of the base Ab= Length × width = 8×6 = 48ft^2

Area of the two triangles At = 2 × 0.5 × width × height

At = width × height = 6×4 = 24ft^2

Area of the two rectangles Ar = 2 × side measurement × length

Side measurement is given as 5ft.

or it can be calculated using;

Side length can be calculated using Pythagoras theorem.

Examining the figure attached, we can see that;

Side breadth b = √(h^2 + (w/2)^2)

b = √(4^2 + (6/2)^2) = √(16+9) = √25

b = 5ft

Ar = 2×5×8 = 80ft^2

From eqn1, S.A can be solved as;

S.A = 48 + 24 + 80

S.A = 152ft^2

7 0
4 years ago
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
4 years ago
A farmer has 624 eggs. He stores them in boxes of 12. How many boxes will he fill?
prisoha [69]
624/12 = 52
He will fill 52 boxes
3 0
3 years ago
Which point is a solution to y&lt;_3x-4?
GenaCL600 [577]

Answer:

D

Step-by-step explanation:

Replace by the coordinates

3*3 -1>1

3 0
3 years ago
Read 2 more answers
X/6 = 8 what is the answer
Dovator [93]

Answer:

x = 48

Step-by-step explanation:

Multiply both sides by 6

x = 6 * 8

x = 48

Check:

48/6 = 8

8 = 8

8 0
3 years ago
Read 2 more answers
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