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3 years ago

13. If l is parallel to m, find the values of x and y.

Mathematics

1 answer:

Delicious77 [7]3 years ago

6 0

Answer: 14. x = 16; y = 23

15. x = 9; y = 13

Step-by-step explanation:

14. (4x + 4) = (7x - 44) (alternate exterior angles are congruent)

4x + 4 = 7x - 44

Collect like terms

4x - 7x = -4 - 44

-3x = -48

Divide both sides by -3

x = -48/-3

x = 16

39° + (8y - 43)° = 180° (consecutive exterior angles are supplementary)

39 + 8y - 43 = 180

Add like terms

-4 + 8y = 180

Add 4 to both sides

8y = 180 + 4

8y = 184

Divide both sides by 8

y = 184/8

y = 23

15. (15x - 26)° = (12x + 1)° (alternate exterior angles are congruent)

15x - 26 = 12x + 1

Collect like terms

15x - 12x = 26 + 1

3x = 27

Divide both sides by 3

x = 27/3

x = 9

28° + (12x + 1)° + (4y - 9)° = 180° (sum of interior angles of ∆)

Plug in the value of x

28 + 12(9) + 1 + 4y - 9 = 180

28 + 108 + 1 + 4y - 9 = 180

Add like terms

128 + 4y = 180

Subtract 128 from each side of the equation

4y = 180 - 128

4y = 52

Divide both sides by 4

y = 52/4

y = 13

Both answer and explantion.

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Read 2 more answers

Rationalize the denominator of $\frac{5}{2+\sqrt{6}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, where $A$, $B$, $C$

horrorfan [7]

Answer:

$A +B+C+D = 3$ is the correct answer.

Step-by-step explanation:

Given:

$\frac{5}{2+\sqrt{6}}$

To find:

$A+B+C+D = ?$ if given term is written as following:

$\frac{A\sqrt{B}+C}{D}$

Solution:

We can see that the resulting expression does not contain anything under $\sqrt$ (square root) so we need to rationalize the denominator to remove the square root from denominator.

The rule to rationalize is:

Any term having square root term in the denominator, multiply and divide with the expression by changing the sign of square root term of the denominator.

Applying this rule to rationalize the given expression:

$\dfrac{5}{2+\sqrt{6}} \times \dfrac{2-\sqrt6}{2-\sqrt6}\\\Rightarrow \dfrac{5 \times (2-\sqrt6)}{(2+\sqrt{6}) \times (2-\sqrt6)} \\\Rightarrow \dfrac{10-5\sqrt6}{2^2-(\sqrt6)^2}\ \ \ \ \ (\because \bold{(a+b)(a-b)=a^2-b^2})\\\Rightarrow \dfrac{10-5\sqrt6}{4-6}\\\Rightarrow \dfrac{10-5\sqrt6}{-2}\\\Rightarrow \dfrac{-5\sqrt6+10}{-2}\\\Rightarrow \dfrac{5\sqrt6-10}{2}$