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3 years ago

Solve each equation. Round your answers to the nearest ten-thousandth. Please show work. Part 3A

Mathematics

1 answer:

Butoxors [25]3 years ago

5 0

Answer:

19.

log9(5x^2 + 10) - log9(10) = 1

<=> log9((5x^2 + 10)/10) = log9(9)

<=> (5x^2 + 10)/10 = 9

<=> 5x^2 + 10 = 90

<=> 5x^2 = 80

<=> x^2 = 16

<=> x = +/- (4)

20.

log5(2x^2 + 4) + log5(3) = 2

<=> log5((2x^2 + 4) x 3) = log3(9)

<=> 6x^2 + 12 = 9

<=> 6x^2 = -3

=> No real x satisfies. ( x^2 always larger or equal to 0)

21.

log6(8) + log6(7 - 2x^2) = 2

<=> log6(8 x (7 - 2x^2)) = log6(36)

<=> 56 - 16x^2 = 36

<=> 16x^2 = 20

<=> x^2 = 5/4

<=> x = +/- sqrt(5/4)

Hope this helps!

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3 years ago

Mitchell is hired to paint a mural on a wall with an area of 21 1/8 square yards. what is the length of the wall?

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3 years ago

The area of a triangular flower bed in the park has an area of

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Answer:

b = 24 ft

h = 10 ft

Step-by-step explanation:

120 = ½hb

120 = ½h(3h - 6)

240 = h(3h - 6)

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8 0

2 years ago

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3t^{2} - 2t - 27 = 0$

$a = 3, b = -2, c = -27$

Then

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328$

$t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35$

$t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685$

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

$y = t^{2} - 4 = (3.35)^2 - 4 = 7.22$

The first point is (27.55, 7.22)

t = -2.685

$x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3$

$y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21$

The second point is (-11.3, 3.21).

8 0

3 years ago

Solve each equation. Round your answers to the nearest ten-thousandth. Please show work. Part 3A​

Solve each equation. Round your answers to the nearest ten-thousandth. Please show work. Part 3A