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Serggg [28]
2 years ago
6

Which equation can be used to find the length of AC?

Mathematics
2 answers:
Ivenika [448]2 years ago
6 0

Answer:

AC=10\sin40^{\circ}

Option 1 is correct.

Step-by-step explanation:

Given the right angled triangle ABC in which right angle is at C

we have to find the length of AC

Hypotenuse=AB=10 in

Base=BC=a

Perpendicular=AC=b

we have to find the equation which can be used to find the length of AC.

Here AC is perpendicular and hypotenuse is known therefore we use sine ratio.

\sin \angle B=\frac{Perpendicular}{Hypotenuse}=\frac{AC}{AB}

\sin40^{\circ}=\frac{AC}{10}

AC=10\sin40^{\circ}

Hence, option 1 is correct.

sladkih [1.3K]2 years ago
5 0

yep, the answer is the first option

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Step-by-step explanation:

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Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate. 5x^6=30
Elina [12.6K]

Answer:

The real solutions are

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Step-by-step explanation:

The solution, or root, of an equation is any value or set of values that can be substituted into the equation to make it a true statement.

To find the real solutions of the equation 5x^6=30:

\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x^6}{5}=\frac{30}{5}\\\\\mathrm{Simplify}\\\\x^6=6\\\\\mathrm{For\:}x^n=f\left(a\right)\mathrm{,\:n\:is\:even,\:the\:solutions\:are\:}x=\sqrt[n]{f\left(a\right)},\:-\sqrt[n]{f\left(a\right)}\\\\x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35

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2 years ago
Please help whoever answers this correctly I'll mark your answer brainliest​
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Answers:

  1. Incorrect
  2. Correct
  3. Correct

==================================================

Explanation:

When applying any kind of reflections, the parallel sides will stay parallel. Check out the diagram below for an example of this.

So PQ stays parallel to RS. Also, QR stays parallel to PS.

The statement "PQ is parallel to PS" is incorrect because the two segments intersect at point P. This letter "P" is found in "PQ" and "PS" to show the common point of intersection. Parallel lines never intersect.

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2 years ago
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