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Anastaziya [24]
3 years ago
6

With a certain tail wind a jet aircraft arrives at its destination, 1,890 miles away, in 3 hours. Flying against the same wind,

the plane makes the return trip in 3 3/8 hours. Find the wind speed and the plane's airspeed.
Mathematics
1 answer:
Gelneren [198K]3 years ago
3 0
<span>Find the wind speed and the plane's airspeed.
:
Let s = speed of the plane in still air
Let w = speed of the wind
then
(s-w) = plane speed against the wind
and
(s+w) = plane speed with the wind
:
Change 3 3/8 hrs to 3.375 hrs
:
The trips there and back are equal distance, (1890 mi) write two distance equations
dist = time * speed
:
3.375(s-w) = 1890
3.0(s + w) = 1890

:
It is convenient that we can simplify both these equations:
divide the 1st by 3.375
divide the 2nd by 3
resulting in two simple equations that can be used for elimination of w
s - w = 560
s + w = 630
----------------adding eliminates w, find s
2s = 1190
s = 
s = 595 mph is the plane speed in still air
Find w
595 + w = 630
w = 630 - 595
w = 35 mph is the wind spee</span>
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A can of beans has surface area 337 cm2. It's heights is 11cm. What is the radius of the circular top?
Karolina [17]

Answer:

3.66 cm

Step-by-step explanation:

r = Radius of cylinder

h = Height of the cylinder = 11 cm

A = Total surface area of the cylinder = 337\ \text{cm}^2

Total surface area of a cylinder is given by

A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi r(r+h)\\\Rightarrow \dfrac{A}{2\pi}=r^2+rh\\\Rightarrow r^2+rh-\dfrac{A}{2\pi}=0\\\Rightarrow r^2+11r-\dfrac{337}{2\pi}=0\\\Rightarrow r^2+11r-53.64=0\\\Rightarrow 100r^2+1100r-5364=0\\\Rightarrow r=\dfrac{-1100\pm \sqrt{1100^2-4\times 100\left(-5364\right)}}{2\times 100}\\\Rightarrow r=3.66,-14.65

So, the radius of the circular top is 3.66 cm.

6 0
2 years ago
CAN SOMEONE PLS HELP ILL GIVE YOU BRAINLIST!! I DONT UNDERSTAND IT .
Harman [31]

Answer:

m∠P = 82°

m∠Q = 49°

m∠R = 49°

Step-by-step explanation:

<em>In the isosceles triangle, the base angles are equal in measures</em>

In Δ PQR

∵ PQ = PR

∴ Δ PQR is an isosceles triangle

∵ ∠Q and ∠R are the base angles

→ By using the fact above

∴ m∠Q = m∠R

∵ m∠Q = (3x + 25)°

∵ m∠R = (2x + 33)°

→ Equate them

∴ 3x + 25 = 2x + 33

→ Subtract 2x from both sides

∵ 3x - 2x + 25 = 2x - 2x + 33

∴ x + 25 = 33

→ Subtract 25 from both sides

∵ x + 25 - 25 = 33 - 25

∴ x = 8

→ Substitute the value of x in the measures of angles Q and R

∵ m∠Q = 3(8) + 25 = 24 + 25

∴ m∠Q = 49°

∵ m∠R = 2(8) + 33 = 16 + 33

∴ m∠R = 49°

∵ The sum of the measures of the interior angles of a Δ is 180°

∴ m∠P + m∠Q + m∠R = 180°

→ Substitute the measures of angles Q and R

∵ m∠P + 49 + 49 = 180

∴ m∠P + 98 = 180

→ Subtract 98 from both sides

∵ m∠P + 98 - 98 = 180 - 98

∴ m∠P = 82°

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