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Katyanochek1 [597]
1 year ago
10

A DC10 airplane travels 3000 km with a tailwind in 3 hr. It travels 3000 km with a headwind in 4 hr. Find the speed of the plane

and the speed of the wind.
Mathematics
1 answer:
svp [43]1 year ago
6 0

The speed of the plane will be 875km/h and the speed of the wind will be 125km/h.

<h3>How to calculate the speed?</h3>

It should be noted that the speed is calculated as:

= Distance/Time

Based on the information, the DC10 airplane travels 3000 km with a tailwind in 3 hr. It travels 3000 km with a headwind in 4 hr.

Therefore, (v + w) = 3000/3

v + w = 1000 .... I

(v - w) = 3000/4

v - w = 750 .... ii

We'll then add both equations together. Therefore, the speed of the plane will be 875km/h and the speed of the wind will be 125km/h.

Learn more about speed on:

brainly.com/question/13740894

#SPJ1

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Two linear functions are shown. Which function has the greater initial​ value?
tangare [24]

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Function A has the greater initial value because the initial value for Function A is 6 and the initial value for Function B is 3.

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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

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3 years ago
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