13. Describe the number of solutions for the equation. (1 point)

Mathematics

1 answer:

Anna11 [10]3 years ago

4 0

Answer:

no solution

Step-by-step explanation:

Given

3(x - 3) = 3x , that is

3x - 9 = 3x ( add 9 to both sides )

3x = 3x + 9 ( subtract 3x from both sides )

0 = 9 ← not possible

This indicates the equation has no solution

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What are the solutions of the equation (2x + 3)^2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve.

iris [78.8K]

U=2x+3, x=u/2-3/2
u^2+8u+11=0
u1=(-8+sqrt(64-44))/2=-4+sqrt(5)
u2=-4-sqrt(5)
x1=u1/2-3/2=-7/2+sqrt(5)/2
x2=u2-3/2=-7/2-sqrt(5)/2

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3 years ago

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A water trough is 7 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 c

valentina_108 [34]

Answer:

The degree of fastness by which the water is rising is 210 seconds

Step-by-step explanation:

The volume of the trough when the water depth is 20 cm is first calculated

Volume of the trough (Trapezoidal Prism) = LH (A + B) × 0.5

Where L is the length of the trough, H is the height of the trough and A and B are parallel width of the top and bottom of the trough

Volume of the trough = 7 × 0.2 (0.3 + 0.7) × 0.5 = 0.7m³

The fastness at which the water is rising is = Volume ÷ water flow rate = 0.7 ÷ 0.2 = 3.5 min = 210 seconds

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2 years ago

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Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .