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3 years ago

What do diffraction and refraction have

Mathematics

1 answer:

nignag [31]3 years ago

4 0

Answer:

They are both due to the wave nature of light.

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Write 25% as a decimal

LuckyWell [14K]

Move the comma 2 time to the left .25

6 0

3 years ago

Read 2 more answers

What is the missing constant term in the perfect square that starts with x2 + 10x ?

kirza4 [7]

Answer:

Okay so I am assuming you mean the third part?

Step-by-step explanation:

Pretty simple. So you see that the varible x^2 does not have a number in front of it, so there is no extra dividing.

A formula for a perfect square is (x^2)+2xy+y^2

Take 10 and divide it by 2.

10/2=5

You have Y. Finding the third part is squaring Y.

5^2=25

Therefore, the third part of the perfect square is 25. If you want to find Y, look at the stuff above.

7 0

4 years ago

In one U.S city, the taxi cost is $2 plus $0.60 per mile. If you are traveling from the airport, there is an additional charge o

anzhelika [568]

Answer:

60 miles

Step-by-step explanation:

Let

x ----> the number of miles

we know that

The number of miles multiplied by $0.60 per mile plus $2 plus the charge of $5.50 for tolls (because you are travelling from the airport) must be equal to $43.50

The linear equation that represent this problem is

$0.60x+2+5.50=43.50$

Solve for x

Combine like terms left side

$0.60x+7.50=43.50$

subtract 7.50 both sides

$0.60x=43.50-7.50$

$0.60x=36$

divide by 0.60 both sides

$x=60\ miles$

5 0

3 years ago

Can somone please help with this, will give brainliest.

Alex777 [14]

Answer: maybe c or d, but I could be wrong :/

Step-by-step explanation:

7 0

3 years ago

Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make

dalvyx [7]

Answer:

Any value of x

Step-by-step explanation:

Given

$f(x) = \sqrt{x^2 - 1}$

$g(x) = \sqrt{x^2 + 1}$

Required

What value of x is $f(g(x)) = g(f(x))$

Solving for f(g(x))

$f(x) = \sqrt{x^2 - 1}$

$f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}$

Solve the inner square

$f(g(x)) = \sqrt{(x^2 + 1 - 1}$

$f(g(x)) = \sqrt{x^2 } }$

$f(g(x)) = x$

Solving g(f(x))

$g(x) = \sqrt{x^2 + 1}$

$g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}$

$g(f(x)) = \sqrt{x^2 - 1 + 1}$

$g(f(x)) = \sqrt{x^2 }$

$g(f(x)) = x$

Equate f(g(x)) and g(f(x))

$f(g(x)) = g(f(x))$

$x = x$

This implies that $f(g(x)) = g(f(x))$ at any value of x

8 0

4 years ago