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3 years ago

The purpose of solving equations is to solve application problems. True or False

2 answers:

5 0

The answer is true, we study mathematics and the properties of numbers, and we learn to solve equations for one major purpose: to solve practical application problems.

Alisiya [41]3 years ago

4 0

The correct answer is True

Explanation:

In mathematics, equations are mathematical statements that include a set of elements that are said to be equal, for example, x + 5 = 4 - 1. These statements are used to find unknown values that are generally represented by letter x, for example by solving the equation previously presented you can find the value of x.

Equations are widely used at school as part of mathematics subject, but besides this, the main purpose of writing and then solving a equation is to find unknown values in real problems or situations either in daily life or other fields such as engineer, for example if you buy a set of items and do not know the value of one of them but know the total price of them you can use an equation to know the unknown value. Thus, it is true the purpose of solving equations is to solve application problems.

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PLEASE HELP!!!! The problem is in the photo below.

Ratling [72]

Answer :36 explanation: (39x39)-(15-15)=1296 then square root that and get 36

7 0

2 years ago

Use function notation to write the equation of the line

blondinia [14]

$f(x)=mx+b\to y=mx+b$

We have two points:

$(-3,\ 0);\ (-4,\ 7)$

The point slope form of the line:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

Substitute

$m=\dfrac{7-0}{-4-(-3)}=\dfrac{7}{-1}=-7$

$y-0=-7(x-(-3))\\\\y=-7(x+3)\\\\y=-7x-21$

Answer:

$f(x)=-7x-21$

8 0

3 years ago

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.

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.

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

3 years ago

A car traveled 88 km in 1 hour, 90 km in the next 2 hours, and then 76 km in 1 hour before reaching its destination. what was th

luda_lava [24]

The formula for Average speed = $\frac{Total Distance}{Total Time}$