Why is it useful to rewrite a formula in terms of another variable?

1 answer:

6 0

Sometimes you are given one value and you have to find another value in a formula. For example if you're pouring a concrete sidewalk you'll need to know the length, width, and depth to calculate the volume of the concrete you'll have to buy. Another possibility is someone gives you concrete and you know the volume of it and you need to find out what size of a sidewalk you can make using the given volume of concrete.

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PLEASE HELP ASAP!!!!

DaniilM [7]

Answer:

An independent clause is a group of words that contains both a subject and a predicate. It expresses a complete thought and can stand alone as a sentence. It can also be joined to other dependent or independent clauses to make a more interesting and complex sentence.Step-by-step explanation:

7 0

2 years ago

Dovator [93]

Answer:

Step-by-step explanation:

5 0

2 years ago

What can be concluded about the line represented in the table? Check all that apply. x y –6 –7 2 –3 8 0 The slope is 2. The slop

Stels [109]

The first thing we must do for this case is to find the equation of the line.
$y-yo = m (x-xo)
$
We have then:
$m = (y2 - y1) / (x2-x1)

m = (-3 - 0) / (2-8)

m = 1/2$
We choose an ordered pair:
$(xo, yo) = (8, 0)
$
Substituting values:
$y-0 = (1/2) (x-8)

y = (1/2) x - 4$

From here we conclude:

Intersection with y:
We evaluate x = 0 in the function:
$y = (1/2) (0) - 4

y = -4$

Slope of the line:
$m = 1/2
$

Point (-2, -5):
We evaluate the value of x = -2 and the value of y = -5
$-5 = (1/2) (- 2) - 4

-5 = -1 - 4

-5 = - 5$
The equation is satisfied.

Point (8, 0):
It is part of the table, therefore belongs to the line.

Answer:
The slope is 1/2
The y-intercept is -4.
The points (-2, -5) and (8, 0) are also on the line.

7 0

2 years ago

Read 2 more answers

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$