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3 years ago

5x+y=27 y=4x using substitution method

2 answers:

7 0

Substitute 4x for y, and solve for x;
5x + (4x) = 27
9x = 27
x = 3
Substitute 3 in for x in the first equation and solve for y;
5(3) + y = 27
15 + y = 27
y = 12

Dvinal [7]3 years ago

4 0

5x + y = 27
substitute: 5x + 4x = 27
combine: 9x = 27
divide (by 9): x = 3

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2 years ago

Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. The tide measures 4.35 ft at

netineya [11]

Answer:

f(x) = 4.35 +3.95·sin(πx/12)

Step-by-step explanation:

For problems of this sort, a sine function is used that is of the form ...

f(x) = A + Bsin(2πx/P)

where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.

It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...

f(x) = 4.35 + 3.95·sin(2πx/24)

The argument of the sine function can be simplified to πx/12, as in the Answer, above.

8 0

3 years ago

Evaluate Functions<br> Given f(x) = 2x + 6 and g(x) = 5x - 1<br> Evaluate f(-2)

ZanzabumX [31]

Answer:

f(-2) = -4x + 6

Step-by-step explanation:

f(x) = 2x + 6

f(-2) = -2(2x) + 6

f(-2) = -4x + 6

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2 years ago

Read 2 more answers

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

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3 years ago