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3 years ago

Find the value of the expression 9.18 · z for z = 2.

Mathematics

1 answer:

loris [4]3 years ago

8 0

Answer: 18.36

Step-by-step explanation:

Replace "z" with "2":

$9.18 \cdot 2 = \boxed{18.36}$ .

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Let a and b be roots of x² - 4x + 2 = 0. find the value of a/b² +b/a²

erastovalidia [21]

Answer:

$\dfrac{a}{b^2}+\dfrac{b}{a^2}=10$

Step-by-step explanation:

Given equation: $x^2-4x+2=0$

The roots of the given quadratic equation are the values of x when $y=0$ .

To find the roots, use the quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=1, \quad b=-4, \quad c=2$

$\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}$

$\textsf{Let }a=2+\sqrt{2}$

$\textsf{Let }b=2-\sqrt{2}$

Therefore:

$\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}$

6 0

2 years ago

Read 2 more answers

Q (x + 40) (3x – 20) R S Find m

xz_007 [3.2K]

Answer:

Step-by-step explanation:

1. x + 40 = 3x - 20

2. 40 = 2x - 20

3. 60 = 2x

4. 30 = x

5 0

3 years ago

Can someone help me with this, thanksssss

Effectus [21]

Answer:17

Step-by-step explanation:Follow pemdas.

Start with parentheses which is (5+2) which gives you 7.

Now next in Pemdas is Exponents. SO do 2 to power of 2 which is 4 .Then do 3 to the power of 3 which is 27.Next add 4 to 27 which is 31.so now do 7 times 2 is 14.Then take 31 and subtarct it from 14 which gives you 17.

Look at pic to see clearly how it was solved and the steps. I color code it to show the steps.