Ask question

Ask question

All categories

Zigmanuir [339]

2 years ago

15

can I please get a labeled homework help im confused and last thing please don't make to much long equation because I cant do so

me of them

1 answer:

zepelin [54]2 years ago

6 0

Answer:

your answer is x >18

Step-by-step explanation:

Jahsiah's mistake was he added 9 to both sides of the equation. He should have subtracted 9 from both sides of the equation.

Here's the solution

-2x+9<-27

-9 -9

-2x/-2x<-36/-2x

when u divide by negative in inequalities, u change the sign. So ur correct answer is x>18

hope this helped.

You might be interested in

In a class of 60 students, 5 recieved a grade of A. What fraction of the class did not recieve a grade of A. Reduce in lowest te

sattari [20]

Ok so if 5 people got an a that means 55 did not receive an a so do 55 over 60 ok they both have a 5 in common 5 goes into 55 11 times and5 goes into 60 12 times so your answer is 11/12

5 0

3 years ago

Read 2 more answers

How do you write 8.11 as an improper fraction

Gre4nikov [31]

8.11 as an improper fraction is 811 over 100 811/100 if you convert

6 0

3 years ago

Solve the following equations:<br> a)<br> 4b = 20

Sidana [21]

B=5
You need to get b by itself on the left side. 4b is multiplying 4 to b. You need to the opposite when you want the variable by itself. You would divide 4 to each side. 4 cancels out on the b so it turns into b. Then 4 divided by 20 is 5. B=5

6 0

2 years ago

Read 2 more answers

Please help xxxxxxxx

Free_Kalibri [48]

Answer:

a. $\bold{45000\times \:10^7}$
b. $\bold{3500}$

Step-by-step explanation:

a. $\left(5\times \:10^3\right)\left(9\times \:10^7\right)$

$5\times 10^3 = 5000\\\\5000\times \:9\times \:10^7$

$\mathrm{Multiply\:the\:numbers:}\:5000\times \:9=45000$

$=45000\times \:10^7$

b. $\left(7\times 10^5\right)\div \left(2\times 10^2\right)$

$\left(7\times 10^5\right)\div \left(2\times 10^2\right)=\frac{7\times \:10^5}{2\times \:10^2}$

$\frac{10^5}{10^2} = 10^3$

$\frac{7\times \:10^3}{2}$

$\mathrm{Factor}\:\: 10^3 : 2^3\times5^3$

$\frac{7\times \:2^3\times \:5^3}{2}$

$\frac{2^3}{2}=2^2$

$7\times \:2^2\times \:5^3=3500$

7 0

2 years ago

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

1 year ago

Read 2 more answers