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Vanyuwa [196]
2 years ago
9

In mathematics, the distance between one point (a) and another point (b), each with coordinates (x,y), can be computed by taking

the differences of their x coordinates and their y coordinates and then squaring those differences. the squares are added and the square root of the resulting sum is taken and... voila! the distance. given two variables, p1 and p2 that are of type point -- a structured type with two fields, x and y, both of type double-- write an expression whose value is the distance between the two points represented by p1 and p

Mathematics
1 answer:
anygoal [31]2 years ago
8 0
The formula to find the distance between points P_{1} and x_{2} is given as \sqrt{ ( y_{1}- y_{2})  ^{2}+ ( x_{1}- x_{2})  ^{2}  }, where

y_{1}- y_{2} is the vertical distance between two points on the y-axis
x_{1} - x_{2} is the horizontal distance between two points on the x-axis

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gladu [14]

Answer:

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2 years ago
Find the value of each variable. Write an<br> equation then solve showing ALL the work<br> A2x+10°
AveGali [126]

Answer:

x = 25

Step-by-step explanation:

The three straight lines in the sides of the triangle means that ALL 3 SIDES ARE EQUAL.

This is an equilateral triangle. So all 3 angles are equal.

We know sum of 3 angles in a triangle is 180. Since 3 of the angles are equal, each angle is:

180/3 = 60

The top angle is given as "2x + 10", thus we can say "2x + 10" is equal to 60 degrees. Now we equate and solve for x:

2x+10=60\\2x=60-10\\2x=50\\x=\frac{50}{2}\\x=25

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4 0
3 years ago
Answer part A, B and part C correctly please.
Radda [10]

Answer:

Part A:

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7 0
3 years ago
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How to solve an expression with variables in the exponents?
gtnhenbr [62]

Answer:

  use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

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You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
2 years ago
Can SOmeone please help me i need to get a good grade on this please!!!!!!
Rudik [331]

Answer:

20 Units

Step-by-step explanation:

Hight times length/by half is tribular prism Formula

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