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2 years ago

PLEASEE answer correctly !!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

natita [175]2 years ago

7 0

Answer:

Step-by-step explanation:

4 boys and 5 girls ratio (4:5)

Anon25 [30]2 years ago

7 0

4:5 ratio

hope this helps

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What is the area of the regular hexagon ABCEF? Explain your reasoning.

Reil [10]

Answer:

1350 $\sqrt{3}$ units²

Step-by-step explanation:

The regular hexagon consists of 6 equiangular triangles

The area (A) of a equilateral triangle is calculated as

A = $\frac{s^2\sqrt{3} }{4}$ ( s is the side length )

Here s = 30 , then

A = $\frac{30^2\sqrt{3} }{4}$ = $\frac{900\sqrt{3} }{4}$ = 225 $\sqrt{3}$ units²

Thus the area of the regular hexagon is

area = 6 × 225 $\sqrt{3}$

= 1350 $\sqrt{3}$ units² ← exact value

≈ 2338.3 units² ( to 1 dec. place )

4 0

2 years ago

Find the area of the figure to the nearest tenth.

Alisiya [41]

Answer:

Step-by-step explanation:

6 x 6 = 36

now to the nearest tenth it would be

so your answer is 36

5 0

3 years ago

Read 2 more answers

Which of the following is a trinomial with a constant term?

Leto [7]

I think the answer is b

4 0

3 years ago

A military cannon is placed at the base of a hill. The cannon is fired at an angle toward the hill. The path

katrin [286]

Answer:

Algebracicaly speaking the answer would be either -13.3876 or - 158.612 through the quadratic equation, but these answers don’t make sense in this real world scenario.

Step-by-step explanation:

7 0

2 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago