What is the distance between 1st and 3rd base?

A number that does not change

Delicious77 [7]

Answer:

A constant

Step-by-step explanation:

Well, ideally I'd like more details, but a number that doesn't change in a mathematical equation is called a constant.

6 0

2 years ago

Read 2 more answers

A website had 2,135,789 hits .what is the value of the digit 3

Hatshy [7]

It is on the ten thousands value

7 0

2 years ago

Simplify the rational expression. State any restrictions on the variable.

ExtremeBDS [4]

we are given

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}$

Firstly, we will factor numerators and denominators

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}$

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

$n^2-6\neq 0$

$n\neq -+ \sqrt{6}$

we can simplify it

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}$

we know that denominator can not be zero

$n^2-3\neq 0$

$n\neq -+ \sqrt{3}$

so, option-B.......Answer

3 0

3 years ago

Questions are in the picture

gtnhenbr [62]

The closest point is (3.5, 1.9) and the distance is 1.96 units

<h3>How to determine the point and the distance?</h3>

The coordinate is given as:

(4, 0)

The equation of the function is

y = √x

The distance between two points is calculated using

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have the following points

(x1, y1) = (4, 0) and (x2, y2) = (x, 0)

This gives

d = √(x - 4)^2 + (√x - 0)^2

Evaluate the difference

d = √(x - 4)^2 + (√x)^2

Evaluate the exponent

d = √x^2 - 8x + 16 + x

Evaluate the like terms

d = √x^2 - 7x + 16

Next, we differentiate using a graphing calculator

d' = (2x - 7)/[2√(x^2 - 7x + 16)]

Set to 0

(2x - 7)/[2√(x^2 - 7x + 16)] = 0

Cross multiply

2x - 7 = 0

Add 7 to both sides

2x = 7

Divide by 2

x = 3.5

So, we have:

Substitute x = 3.5 in y = √x

y = √3.5

Evaluate

y = 1.9

So, the point is (3.5, 1.9)

The distance is then calculated as:

d = √(x2 - x1)^2 + (y2 - y1)^2

This gives

d = √(3.5 - 4)^2 + (1.9 - 0)^2

Evaluate

d = 1.96

Hence, the closest point is (3.5, 1.9) and the distance is 1.96 units