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3 years ago

How do you simplify quadratic equations?

Mathematics

1 answer:

Scrat [10]3 years ago

8 0

Answer:

To solve a quadratic equation by factoring,

Put all terms on one side of the equal sign, leaving zero on the other side.

Factor.

Set each factor equal to zero.

Solve each of these equations.

Check by inserting your answer in the original equation.

Step-by-step explanation:

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The function f(x) is graphed on the coordinate plane.

tatiyna

F(x) = 1/4x-1

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6 0

3 years ago

Read 2 more answers

Identify the solution set of 3 In 4 = 2 In x.<br> -{6}<br> -{-8,8}<br> -(8)

o-na [289]

Option C: The solution set is $\{8\}$

Explanation:

The expression is $3 \ln 4=2 \ln x$

Now, let us find the solution set.

Switch sides, we get,

$2 \ln x=3 \ln 4$

Dividing by 2 on both sides, we have,

$\ln x=\frac{3 \ln 4}{2}$

Thus, $\ln 4=\ln 2^2=2\ln 2$

Hence, the above expression becomes,

$\ln x=\frac{3 \cdot 2 \ln (2)}{2}$

Simplifying, we get,

$\ln x=3 \ln 2$

Applying the log rule, we get,

$$\ln x$=\ln \left(2^{3}\right)$

Simplifying, we have,

$$\ln x$=\ln \left8\right$

Applying the log rule, we have,

$x=8$

Thus, the solution set is $\{8\}$

Hence, Option C is the correct answer.

5 0

3 years ago

√3=1.732 find the value of√75+1/2√48-√198

Nookie1986 [14]

Answer:

12.124-3 $\sqrt{22}$

Step-by-step explanation:

$\sqrt{75} = \sqrt{5^{2}*3} \\$

$\sqrt{48}= \sqrt{2^{2}*2^{2}*3} \\$

$\sqrt{198} =\sqrt{3^{2}*2*11} \\$

$\sqrt{75} +\frac{1}{2} *\sqrt{48} - \sqrt{198} \\\\\sqrt{5^{2}*3} +\frac{1}{2} *\sqrt{2^{2}*2^{2}*3} - \sqrt{3^{2}*2*11} \\\\5\sqrt{3} +\frac{1}{2} *4 \sqrt{3} - 3 \sqrt{2*11} \\\\5\sqrt{3} +2 \sqrt{3} - 3 \sqrt{22} \\\\5*1.732 +2*1.732 - 3 \sqrt{22} \\\\8.66 + 3.464 - 3 \sqrt{22}\\\\12.124- 3 \sqrt{22}$