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3 years ago

These I need help on

Mathematics

1 answer:

Luba_88 [7]3 years ago

8 0

Answer:

I did the first page you asked for just now! :)

Step-by-step explanation:

20) 15 = 3r R = 5

21) k/2 = 6 K = 12

22) 3h = 27 H = 9

23) j/3 = 4 J = 12

24) 6p = 30 P = 5

25) n/10 = 40 N = 400

26) h/4 = 15 H = 60

27) 9/s = 81 S = 9

28) 14 = 2c C = 7

29) 26 = 13d D = 2

30) 6 = m/3 M = 18

31) 7 = p/5 P= 35

32) 4w = 16 W = 4

33) f/3 = 9 F = 27

34) 20 = 4x X = 5

35) 3z = 36 Z = 12

36) 10 = j/6 J = 60

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Answer:

$\large\boxed{x=\dfrac{9}{8}-\dfrac{\sqrt{109}}{8},\ y=-\dfrac{1}{2}-\dfrac{\sqrt{109}}{18}}\\or\\\boxed{x=\dfrac{9}{8}+\dfrac{\sqrt{109}}{2},\ y=-\dfrac{1}{2}+\dfrac{\sqrt{109}}{18}}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x\times9y=7&(1)\\4x-9y=9&(2)\end{array}\right\\\\(2)\\4x-9y=9\qquad\text{subtract}\ 4x\ \text{from both sides}\\-9y=-4x+9\qquad\text{change the signs}\\9y=4x-9\qquad\text{substitute it to (1)}\\\\4x(4x-9)=7\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\(4x)(4x)+(4x)(-9)=7\\(4x)^2-36x=7\\(4x)^2-2(4x)(4.5)=7\qquad\text{add}\ 4.5^2\ \text{to both sides}\\(4x)^2-2(4x)(4.5)+4.5^2=7+4.5^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$(4x-4.5)^2=7+20.25\\(4x-4.5)=27.25\to 4x-4.5=\pm\sqrt{27.25}\\\\4x-\dfrac{45}{10}=\pm\sqrt{\dfrac{2725}{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{2725}}{\sqrt{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25\cdot109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{5\sqrt{109}}{10}\qquad\text{add}\ \dfrac{45}{10}\ \text{to both sides}\\\\4x=\dfrac{45}{10}\pm\dfrac{5\sqrt{109}}{10}$

$4x=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 4}\\\\x=\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\\\\\text{Put the values of}\ x\ \text{to (2):}\\\\9y=4\left(\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\right)-9\\\\9y=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}-\dfrac{18}{2}\\\\9y=-\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 9}\\\\y=-\dfrac{1}{2}\pm\dfrac{\sqrt{109}}{18}$