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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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Sandy evaluated the expression below

andre [41]

Answer:

Sandy should have evaluated (-2)³ as -8

Step-by-step explanation:

(-2)³ (6 - 3) - 5(2 + 3) = -8 · 3 - 5 · 5 = - 24 - 25 = - 49

Because:

(-2) · (-2) · (-2) = 4 · (-2) = - 8

God with you!!!

8 0

3 years ago

Read 2 more answers

The graph 4x^2-4x-1 is shown. Use the grpah to find the estimates for the solutions of 4x^2-4x-1=0 and 4x^2 - 4x-1=2

Darina [25.2K]

Answer:

a) The estimates for the solutions of $4\cdot x^{2}-4\cdot x -1 = 0$ are $x_{1}\approx -0.25$ and $x_{2} \approx 1.25$ .

b) The estimates for the solutions of $4\cdot x^{2}-4\cdot x -1 = 2$ are $x_{1}\approx -0.5$ and $x_{2} \approx 1.5$

Step-by-step explanation:

From image we get a graphical representation of the second-order polynomial $y = 4\cdot x^{2}-4\cdot x -1$ , where $x$ is related to the horizontal axis of the Cartesian plane, whereas $y$ is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case: