Solve for x. The polygons are similar

Mathematics

1 answer:

kotykmax [81]3 years ago

4 0

$\frac{8x-2}{42}=\frac{63}{49}\\\frac{8x-2}{42}=\frac{9}{7}\\7(8x-2)=9\cdot42\ /:7\\8x-2=9\cdot6\\8x-2=54\\8x=54+2\\8x=56\ /:8\\x=7$

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Peter attends 6 lessons each week . A year he attends 52 weeks. But he missed 5 classes how many lessons did he attend during th

VMariaS [17]

You are given the information that he takes six lessons per week. First we will calculate the total lessons he could potentially take per year.

6 • 52 = 312

He could potentially take 312 lessons in one year.

Now that you know this information, you simply subtract the days he missed from that total.

312 - 5 = 307

Your final answer: Peter took 307 lessons during the year.

4 0

3 years ago

Find sin 2a and cot 2a:

geniusboy [140]

Answer:

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cot(2\alpha ) = \frac{16511}{18720}$

Step-by-step explanation:

$\text{if } cos(\alpha)=\frac{12}{13}\\\text{That must mean we have a triangle with base 12, and hypotenuse 13.}\\\text{Using Pythagoras, we can determine the base of the triangle must be 5.}\\a^2+b^2=c^2 \text{, where c is the hypotenuse and a, b are the two other sides.}\\c^2-b^2=a^2\\\sqrt{c^2-b^2}=a\\\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\\\text{Therefore, }sin(\alpha) = \frac{5}{12}\\sin(2\alpha)=2sin(\alpha )cos(\alpha)\\\text{(From double angle formulae identities)}\\$

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cos(2\alpha )=cos^2(\alpha)-sin^2(\alpha)\\cos(2\alpha )=(\frac{12}{13})^2-(\frac{5}{12})^2=\frac{16511}{24336}\\cot(2\alpha)=\frac{cos(2\alpha)}{sin(2\alpha)}=\frac{\frac{16511}{24336}}{\frac{10}{13}}=\frac{16511}{18720}$