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IrinaK [193]
3 years ago
11

The base and height of the triangular mainsail of a sailboat are 1.5 times greater than the base and height of the boat's triang

ular staysail. How many times greater is the area of the mainsail than the area of the staysail?
Mathematics
1 answer:
nataly862011 [7]3 years ago
4 0
Let say the mainsail length and width is L1 and W1, staysail length and width is L2 and W2. 
<span>The base and height of the triangular mainsail of a sailboat are 1.5 times greater than the base and height of the boat's triangular staysail. That mean:
L1 = 1.5L2
W1 =1.5 W2

</span>How many times greater is the area of the mainsail than the area of the staysail?area1/area2 =
L1*W1 / L2*W2 =
1.5L2 * 1.5W2/ L2*W2 = 1.5 * 1.5= 2.25 times
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The seating capacity of a theater is 250. Tickets are $3 for children and $5 for adults. The theater must take in at least $1100
maksim [4K]

Answer:

x + y ≤

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Step-by-step explanation:

Given:

Seating capacity of theater = 250

Cost of each child ticket = $3

Cost of each adult ticket = $5

Cost per performance = $1100 at least

Find:

System of inequalities

Computation:

Let;

x = Number of children's tickets

y = Number of adult tickets

So

x + y ≤

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Which one of the following is a better buy: a large pizza with a 14-inch diameter for $13.00 or a medium pizza with a 7-inch dia
bija089 [108]

Answer:

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(1,10), (18.-10) slope
kobusy [5.1K]

Answer:

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2. The electrical resistance R of a wire varies inversely with the square of its diameter d. If a wire with a
Mariulka [41]

Answer:

0.225 \Omega =225 m\Omega

Step-by-step explanation:

We know (Ohm second law) that R= \frac{k}{d^2} where k inlcudes the rest of the parameters (material, lenght). In our situation we have k= 0.4\cdot9 \Omega mm^2. The moment the diameter becomes 8mm R becomes

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The volume V of a solid right circular cylinder is given by V = πr2h where r is the radius of the cylinder and h is its height.
emmainna [20.7K]

Answer:

8.20in³

Step-by-step explanation:

Given V = πr²h

r is the radius = 1.5in

h is the height = 6in

thickness of wall of the cylinder dr = 0.04in

top and bottom thickness dh 0.07in+0.07in = 0.14in

To compute the volume, we will find the value of dV

dV = dV/dr • dr + dV/dh • dh

dV/dr = 2πrh

dV/dh = πr²

dV = 2πrh dr + πr² dh

Substituting the values into the formula

dV = 2π(1.5)(6)•(0.04) + π(1.5)²(6) • 0.14

dV = 2π (0.36)+π(1.89)

dV = 0.72π+1.89π

dV = 2.61π

dV = 2.61(3.14)

dV = 8.1954in³

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