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Montano1993 [528]
3 years ago
10

Abigail is 8 years older than Cynthia. Twenty years ago Abigail was three times as old as Cynthia

Mathematics
1 answer:
Anestetic [448]3 years ago
5 0
Abigail would be 32 and Cynthia would be 24 now. Twenty years ago, Abigail was 12, and Cynthia was 4. 12 is 3 times as much as 4, and also 4+8=12.
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Find the area 20 points!!!
Dmitriy789 [7]

Answer:

the answer would be 29 :)

Step-by-step explanation:

5 0
3 years ago
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Country days scholarship rounds receive a gift of $135000. The money is invested in stock, bonds, and CDs. CDs pay 2.75% interes
Vsevolod [243]

Answer:

  • CDs — $10,000
  • bonds — $80,000
  • stocks — $45,000

Step-by-step explanation:

Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:

  c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested

  0.0275c +.045b +0.104s = 8555 . . . . . total income earned

  -c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs

Using the third equation to write an expression for b, we can substitute into the other two equations.

  b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b

  c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation

  2c +s = 65000 . . . . . . . . . . . . . . . . [eq4] simplify

  .0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn

  .0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify

Using [eq4], we can write an expression for s that can be substituted into [eq5].

  s = 65000 -2c . . . . . . . expression we can substitute for s

  0.0725c +0.104(65000 -2c) = 5405

  -0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify

  c = 1355/.1355 = 10,000

  s = 65000 -2×10000 = 45,000

  b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]

6 0
3 years ago
What is the solutions to 0=x^2-x-6<br> two things <br> ,,quadratic equationz,,,
natita [175]

Answer:

x1=3. x2=-2

Step-by-step explanation:

x =  \frac{ - b +  -  \sqrt{b {}^{2} - 4ac } }{2a}

b=-1

a=1

c=-6

x =  \frac{ - ( - 1) +  -  \sqrt{ {( - 1)}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1}

x =  \frac{1 +  -  \sqrt{1 + 24} }{2}

x =  \frac{1 + -   \sqrt{25} }{2}

x1 =  \frac{1 + 5}{2}

x2 =  \frac{1 - 5}{2}

x1=3

x2=-2

x =  \frac{1 +  - 5}{2}

3 0
3 years ago
(X+8)(x-8) if necessary combine like terms
julia-pushkina [17]

Answer:

x² - 64

Step-by-step explanation:

Given

(x + 8)(x - 8)

Each term in the second factor is multiplied by each term in the first factor, that is

x(x - 8) + 8(x - 8) ← distribute both parenthesis

= x² - 8x + 8x - 64 ← collect like terms

= x² - 64

3 0
3 years ago
Desmond wants to sell his car that he paid $8,000 for 2 years ago. The car depreciated, or decreased in value, at a constant rat
butalik [34]

Answer:

8,000-24x

Step-by-step explanation:

there's 24 months in 2 years. If it's value decreased from 8,000 at a constant rate over a 2 year period, the equation would be 8,000-24x

5 0
3 years ago
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