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3 years ago

Round 0.978 to the nearest tenth

1 answer:

5 0

The answer is one, because the number seven in the hundredths column is greater than five so you round the nine up to 1.0.

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Write equation in standard form: 1.y=3/4x+5 2.y+4=1/3(x+6)

Gwar [14]

1. 3x - 4y = 20

2. x - 3y = -6

5 0

3 years ago

kondaur [170]

Answer:

Karim will never have triple age as his brother, it has happened already 1 year ago, thus he won't give him a gift

Step-by-step explanation:

Equations

Let's make:

x = the number of years from now

Karim now is 22 and Ziad is 8. x years from now Karim will be

22 + x years old and Ziad will be

8 + x years old.

Karim will give Ziad a wonderful gift when:

22 + x = 3(8 + x)

22 + x = 24 + 3x

Solving:

x = -1

This means that Karim will never have triple age as his brother, it has happened already 1 year ago, thus he won't give him a gift

4 0

3 years ago

Determine the commutators of the operators a and a+,where a = (x + ip)/2 ^1/2 and a+ = (x - ip)/ 2 ^1/2

Vsevolod [243]

Answer:

Given that:

$a = (\frac{x+ip}{2})^{\frac{1}{2}}$ and $a+= (\frac{x-ip}{2})^{\frac{1}{2}}$

if a , a+ commutator, it obeys $aa^+ = a^+a$

First find:

$aa^+ = (\frac{x+ip}{2})^{\frac{1}{2}} (\frac{x-ip}{2})^{\frac{1}{2}}$

= $(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}$

Now;

$a^+a =(\frac{x-ip}{2})^{\frac{1}{2}} (\frac{x+ip}{2})^{\frac{1}{2}} = (\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}$

= $(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}$

therefore, $aa^+ = a^+a$ which implies the operators a and a+ are commutators.

7 0

3 years ago

F(x)=<img src="https://tex.z-dn.net/?f=%28x%2B1-%7Cx-1%7C%29%2F2" id="TexFormula1" title="(x+1-|x-1|)/2" alt="(x+1-|x-1|)/2" ali

katrin [286]

Answer:

$f(g(x))=\dfrac{|x|+1-||x|-1|}{2}$

Step-by-step explanation:

The most straightforward way is direct substitution of g(x) for x in f(x):

$f(g(x))=\dfrac{|x|+1-||x|-1|}{2}$

8 0

3 years ago

Use the long division method

sukhopar [10]

Step-by-step explanation:

please mark me as brainlest

4 0

2 years ago