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Alex73 [517]
3 years ago
15

Khan academy how to solve equations with fractions and mixed numbers

Mathematics
1 answer:
Rufina [12.5K]3 years ago
8 0
Generally, you are told to approach these by "clearing fractions". That is, you generally multiply the equations by the least common denominator so all fractions and mixed numbers become integers.

Alternatively, you can simply do the arithmetic using the numbers given. You learned a long time ago how to add, subtract, multiply, and divide mixed numbers and fractions. Do these operations as necessary to solve the equations.
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3 years ago
In 2000, the population of Big Springs was 13 thousand. Use the given doubling
trasher [3.6K]

Answer:

The answer is "26179.4".

Step-by-step explanation:

Assume year 2000 as t, that is  t =0.

Formula:

A= A_0e^{rt}

Where,

A_0 = \ initial \ pop \\\\r= \ rate \ in \ decimal \\\\t= \ time \ in \ year

for doubling time,

r = \frac{log (2)}{t} \\

r = \frac{\log (2)}{ 40} \\\\r= \frac{0.301}{40}\\\\r= 0.007

Given value:

A = A_0e^{rt} \\\\

A_0 = 13000

t= 40 \ years

when year is 2000, t=0 so, year is 2100 year as t = 100.

A = 13000 \times e^{et}\\\\A = 13000 \times e^{e \times t}\\\\A = 13000 \times e^{0.007 \times 100}\\\\A = 13000 \times e^{0.7}\\\\A= 13000\times 2.0138\\\\A = 26179.4

7 0
3 years ago
Someone help me please
prisoha [69]
The answer is five at least i hope it is

5 0
3 years ago
Read 2 more answers
A The length of a rectangle is 4 m more
Lady bird [3.3K]

Given :

  • The length of a rectangle is 4m more than the width.
  • The area of the rectangle is 45m²

⠀

To Find :

  • The length and width of the rectangle.

⠀

Solution :

We know that,

\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:

So,

Let's assume the length of the rectangle as x and the width will be (x – 4).

⠀

Now, Substituting the given values in the formula :

\qquad \sf \: { \dashrightarrow x  \times  (x - 4) = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x - 45 = 0 }

\qquad \sf \: { \dashrightarrow {x}^{2}    - 9x+ 5 x - 45 = 0 }

\qquad \sf \: { \dashrightarrow x(x - 9) + 5(x - 9) = 0 }

\qquad \sf \: { \dashrightarrow (x  - 9) (x  + 5) = 0 }

\qquad \sf \: { \dashrightarrow x = 9, \: \: x =  - 5}

⠀

Since, The length can't be negative, so the length will be 9 which is positive.

⠀

\qquad { \pmb{ \bf{ Length _{(rectangle)} = 9\:m}}}\:

\qquad { \pmb{ \bf{ Width _{(rectangle)} = 9 - 4=5\: m}}}\:

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8 0
2 years ago
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