All categories

2 years ago

What is the weight of the plane with 81 gallons of fuel in its tanks?

Mathematics

1 answer:

JulijaS [17]2 years ago

3 0

The weight of a plane with 81 gallons of fuel in the tank is about 44 tons

You might be interested in

Solve 5 tan x = 5√(3) for 0° ≤ x ≤ 180°. a. 60° b. 150° c. 30° d. 120°

ladessa [460]

Answer:

$\boxed{a.\:\:\:60\degree}$

Step-by-step explanation:

The given trigonometric equation is

$5\tan(x)=5\sqrt{3}$ for $0\degree \le x\le 180\degree$ .

We divide through by 5 to get;

$\tan(x)=\sqrt{3}$

We take the inverse tangent of both sides to obtain;

$x=tan^{-1}(\sqrt{3})$

$x=60\degree$

Since the tangent ratio is positive, it means $x$ could also be in the 3rd quadrant. But this is outside the given range.

Therefore the correct answer is A

6 0

3 years ago

What's 21/50 in simplest form

andreev551 [17]

21/50

21/50 is in its most simple form because it cannot be broken down any further.

Answer: 21/50

5 0

3 years ago

Read 2 more answers

4. The school spirit club is stuffing bags for

Dima020 [189]

Answer:

2.222

Step-by-step explanation:

there is 1000 pieces of candy and 450 bags so 1000 ÷ 450 = 2.222

8 0

2 years ago

If you are givent two points on a line, (5,6) and (-5, -4), what is the slope of the line!

alexira [117]

Answer:

m = -1

Step-by-step explanation:

Formula we'll use: $\frac{y2}{x2} -\frac{y1}{x1} =m$

m = slope.

y2 = -4

x2 = -5

y1 = 6

x1 = 5

Substitute -4, -5, 6, 5 into the formula:

$\frac{-4}{-5} -\frac{6}{5} =$

Subtract both top and bottom:

m = -10/-10

m = -1

3 0

1 year ago

In △ABC, point D∈ AC with AD:DC=4:3, point E∈ BC so that BE:EC=1:5. If ACDE=5 in2, find ABDC, AABD, and AABC.

Kaylis [27]

Answer:

ar ΔBDC=6 $in^2$

ar ΔABD=8 $in^2$

ar ΔABC=14 $in^2$

Explanation:

Please take a look of attach figure.

We are given area of triangle ar ΔCDE=5 $in^2$

ar ΔCDE= $\frac{1}{2}\times DN\times EC$ ------------(1)

ar ΔBDE= $\frac{1}{2}\times DN\times BE$ ------------(2)

Divide eq(2) by eq(1) and we get,

$\frac{ar\ ΔBDE}{ar\ ΔCDE}=\frac{BE}{EC}$

$\frac{ar\ ΔBDE}{5}=\frac{1}{5}$

So, ar ΔBDE=1 $in^2$

ar ΔBDC=ar ΔBDE+ ar ΔCDE

ar ΔBDC=1+5 = 6 $in^2$

Similarly,

$\frac{ar\ ΔABD}{ar\ ΔBDC}=\frac{AD}{DC}$

$\frac{ar\ ΔABD}{6}=\frac{4}{3}$

ar ΔABD=8 $in^2$

ar ΔABC=ar ΔABD + ar ΔBDC

ar ΔABC=8+6 = 14 $in^2$

6 0

2 years ago