How to solve the problem 3.2d-4d=2.3d+3

1 answer:

5 0

3.2d - 4d = 2.3d + 3...simplify by combining like terms
-0.8d = 2.3d + 3....subtract 2.3d from both sides
-0.8d - 2.3d = 3 ...simplify again
-3.1d = 3...divide both sides by -3.1
d = 3/ -3.1
d = - 0.97

or
3.2d - 4d = 2.3d + 3....multiply the equation by 10, gets rid of the decimals
32d - 40d = 23d + 30....subtract 23d from both sides
32d - 40d - 23d = 30....simplify
-31d = 30...divide by -31
d = -30/31
d = - 0.97

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TiliK225 [7]

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Step-by-step explanation:

5 0

3 years ago

In 2000, the circulation of a local newspaper was 3,250. In 2001, its circulation was 3,640. In 2002, the circulation was 4,100.

vodka [1.7K]

Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.

Solution:

Given that,

Total circulation of local newspaper in 2000 = 3250

Total circulation of local newspaper in 2001 = 3640

Total circulation of local newspaper in 2002 = 4100

Finding percent of increase in the newspaper’s circulation from 2000 to 2001:

$\text { Percent increase from } 2000 \text { to } 2001=\frac{\text { change in circulation from 2001 - 2000 }}{\text { circulation in } 2000} \times 100$

$\begin{array}{l}{=\frac{3640-3250}{3250} \times 100} \\\\ {=\frac{390}{3250} \times 100=12 \%}\end{array}$

Finding percent of increase in the newspaper’s circulation from 2001 to 2002:

$\begin{array}{l}{\text { Percent increase from } 2001 \text { to } 2002=\frac{\text { change in circulation }}{\text {circulation in } 2001} \times 100} \\\\ {=\frac{4100-3640}{3640} \times 100} \\\\ {=\frac{460}{3640} \times 100=12.6 \%}\end{array}$