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3 years ago

Find the zeros of the function F(x)=x^2-5x+5

1 answer:

6 0

You would have to replace f(x) with zero and use quadratic formula

f(x)=x^2-5x+5
0=x^2-5x+5

=(-b+-√b^2-4ac)/2a
=(5+-√25-4(1)(5))/2
=(5+-√5)/2

x=(5+√5)/2 or x=(5-√5)/2

Both x values are completely applicable in the equation.

Hope I helped :)

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What's the square root of 2 rounded to the nearest tenth ?

vichka [17]

It is 1.4 when rounded

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3 years ago

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PLEASE HELP ITS TIMED Order the rational numbers from least to greatest -2/5,-4/3, -1/2, -1

IgorC [24]

Answer:

-2/5, -1/2, -1, -4/3

Step-by-step explanation:

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2 years ago

Please choose an answer below. A) 3.4 B) 36.0 C) 38.6 D) 41.2

schepotkina [342]

Answer:

It would be D.

Step-by-step explanation:

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3 years ago

The sum of two numbers is 60 and the difference is 2 what are the numbers.

oksian1 [2.3K]

Let the two numbers be x and y
Then
x + y = 60
And
x - y = 2
x = y + 2
Putting the value of x in the first equation we get
x + y = 60
y + 2 + y= 60
2y + 2 = 60
2y = 60 - 2
2y = 58
y = 58/2
y = 29
Now putting the value of y in the first equation we get
x + y = 60
x + 29 = 60
x = 60 - 29
x = 31
So the value of the two numbers comes out to be 31 and 29. I hope the procedure is clear enough for you to understand.

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3 years ago

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What is the distance between coordinates (-6.25,5) and (-3.75,2)

Oksi-84 [34.3K]

Answer:

$\sqrt{15.25}$

Step-by-step explanation:

To find the distance, we can use the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ when the given points are $(x_1,y_1)$ and $(x_2,y_2)$

Plug in the points (-6.25,5) and (-3.75,2)

$d=\sqrt{(-6.25-x_1)^2+(5-y_1)^2}\\d=\sqrt{(-6.25-(-3.75))^2+(5-2)^2}\\d=\sqrt{(-6.25+3.75)^2+(5-2)^2}\\d=\sqrt{(-2.5)^2+(3)^2}\\d=\sqrt{6.25+9}\\d=\sqrt{15.25}$

Therefore, the distance between the two coordinates is $\sqrt{15.25}$ , or approximately 3.91.

I hope this helps!

6 0

2 years ago