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3 years ago

Find the imaginary solutions to the following quadratic function:

Mathematics

1 answer:

Temka [501]3 years ago

8 0

Answer:

the solutions are these

Step-by-step explanation:

$\frac{3}{14} + \frac{ \sqrt{159} }{14} i \: and \\ \: \frac{3}{14} - \frac{ \sqrt{159} }{14} i$

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Solve this equation for x. Round your answer to the nearest hundredth. 0.77 = log x

Alekssandra [29.7K]

Answer:

x ≈ 5.89

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra II

Exponential to Logarithmic: $\displaystyle b^m=x \rightarrow log_bx=m$

Step-by-step explanation:

Step 1: Define

Identify

0.77 = log(x)

Step 2: Solve for x

[Equality Property] Raise both sides to the 10th power: $\displaysytle 10^{0.77} = 10^{logx}$
Simplify: $\displaysytle x = 10^{0.77}$
Evaluate: $\displaysytle x = 5.88844$

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Answer: The answer is $44,496. I got it right on the test. Sorry that I was late...

Step-by-step explanation:

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3 years ago

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Dafna11 [192]

Answer: 4 units

Step-by-step explanation:

(5-1,3-3)=(4,0)==> 4 units

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2 years ago

Which statement is true? A. ABD DBC by SSS B. ABD DBC by SAS C. ABE DBC by SAS D. ABE CBD by SSS

Bumek [7]

Answer:

D. ABE =CBD by SSS.

Step-by-step explanation:

Given that AB = BD and BE = BC, the lengths of AE and CD are equal to each other. Hence, triangles ABE and BCD are congruent by SSS Theorem. The answer is D.

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Alex787 [66]

Answer:

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Step-by-step explanation:

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