MATHEMATICS MIDDLE SCHOOL

X^2=14x-49. i just dont know how to do the steps to solve this equation

Answers

Answer 1
Answer: x^2=14x-49 \n \n x^2 - 14x + 49 = 0 \ / \ move \ terms \n \n x^2 - 2(x)(7) + 7^2 = 0 \ / \ rewrite \n \n (x - 7)^2 = 0 \ / \ square \ of \ difference \ method \n \n x - 7 = 0 \ / \ square \ root \ each \ side \n \n x = 7 \ / \ add \ 7 \n \n

The answer is: x = 7.

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Morgan has 1/3 of a pond of bottle caps storend evenly in 3 boxes. How many pounds of bottles caps are in each box?

Liang went to the bookstore. he spent 1/3 of his ,money on a pen and 4/7 of it on books. what fraction of his money did he have left?

Answers

Convert into like fractions --> 
1/3 = 7/21 4/7 = 12/21
7+12 = 19
19/21 is spent therefore he has 2/21 left

The fraction of the amount Liang has left is (2)/(21)

  • Let X = Amount spent on a pen; X = (1)/(3)
  • Let Y = Amount spent on books; Y = (4)/(7)
  • Let M = Amount of money Liang has got; M =?
  • Let L = Amount of money Liang will have left; L = ?

Further Explanation

To derive L (amount of money Liang will have left), add X (amount spent on a pen) and Y (amount spent on books) together, then the derived value be subtracted from M (fraction of total amount Liang has got) i.e.: M – (X + Y)

Therefore, L = M – (X + Y)

Deriving (X + Y)

= X + Y

= (1)/(3) + (4)/(7)

= (1 x 7) + (3 x 4) / 21

= (7) + (12) / 21

= 7 + 12 / 21

= (19)/(21)

X + Y = (19)/(21)

Since the lowest common denominator of the fraction of X and Y = 21, then the fraction of M = (21)/(21)

Thus, M = (21)/(21)

Recall, L = M – (X + Y)

Thus,

L = (21)/(21) – (19)/(21)

L = (1 x 21) – (1 x 19) / 21

L = (21) – (19) /21

L = 21 – 19 /21

L = (2)/(21)

Hence, the amount Liang will have left is (2)/(21)

LEARN MORE:

KEYWORDS:

  • denominator
  • numerator
  • numbers
  • fraction
  • liang

Water is draining out of a 25 gallon cooler at a rate of 3 gallons a minute. How much water is left in the cooler after 5 minutes?

Answers

Minute 1:
25 - 3 = 22

Minute 2:
22 - 3 = 19

Minute 3:
19 - 3 = 16

Minute 4:
16 - 3 = 13

Minute 5:
13 - 3 = 10

After 5 minutes, your cooler has 10 gallons left inside it.
3•5=15
25-15=10

You will have 10 gallons left

What is (-7x + 1) + (-x + 2)

Answers

(-7x + 1) + (-x + 2)=\n-7x+1-x+2=\n-8x+3

How many times larger is 9 x 10^-4 than 3 x 10^-10? A) 3 x 105
B) 3 x 106
C) 6 x 105
D) 6 x 106

Answers

9 x 10^-4: Since the exponent of the scientific notation is negative −4, move the decimal point 4 places to the left. Answer: 0.0009

3 x 10^-10: Since the exponent of the scientific notation is negative −10, move the decimal point 10 places to the left. Answer: 0.0000000003

Subtract: C) 6 x 105

Sam cut 62 and 1/2 inches of ribbon into 4 equal hair ribbon how long was each hair ribbon

Answers

If the total length of ribbon is 62 and 1/2 inches, or 62.5 inches, and he cut it in to 4 equal parts then:

62.5/4 = x 
x = 15.625 inches 

Each ribbon is 15.625 inches.
^^^^^^^^ That Would Be It.. I Think

How do i solve question 25

Answers


I never learned to do operations with absolute values, so I don't have
the standard tools to use.  The only thing I have to work with when I run
into one of these is my brain.  So there are probably better ways to do it,
but here's how I would approach this one:

The fraction itself is either positive or negative.  Either way, its absolute
value is less than 4.

If the fraction is positive, then you can just throw away the absolute-value
bars, and work it like a normal inequality ... multiply each side by (x-3), then
work it down so there's nothing but 'x' on the left side, etc.

If the fraction is negative AND its absolute value is less than 4, then the
fraction itself is greater than -4 .  Throw away the absolute-value bars,
change the inequality to say that, and then work it like a normal inequality.

Do BOTH of these, and you'll come out with two inequalities for 'x' ... it's
more than something AND less than something else.  Put these together
into one inequality, and you have the whole range for what 'x' can be.  

Sounds complicated.  One of these days, I should learn the right way
to do these.
(x+2)/(x-3) \ \textgreater \ 4         or        (x+2)/(x-3) \ \textless \ -4

x+2 \ \textgreater \ 4(x-3)

x+2 \ \textgreater \ 4x-12

x+14 \ \textgreater \ 4x

14 \ \textgreater \ 3x

4 (2)/(3) \ \textgreater \ x  <-answer



(x+2)/(x-3)\ \textless \ - 4

x+2\ \textless \ - 4(x-3)

x+2\ \textless \ - 4x+12

x-10\ \textless \ - 4x

-10\ \textless \ - 5x

2\ \textless \ x <-answer



2 < x < 14/3
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