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777dan777 [17]
3 years ago
8

hi i'm really struggling with this and its due later today! If someone could show the work of how to do it and the answer that w

ould be really appreciated!!

Mathematics
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

\frac{-5 \pm \sqrt{13} }{6}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\frac{-5 \pm \sqrt{5^2-4(3)(1)} }{2(3)}<u />

<u />

<u>Step 2: Evaluate</u>

  1. Evaluate Exponents:                    \frac{-5 \pm \sqrt{25-4(3)(1)} }{2(3)}
  2. Evaluate Multiplication:               \frac{-5 \pm \sqrt{25-12} }{6}
  3. Evaluate Subtraction:                  \frac{-5 \pm \sqrt{13} }{6}
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Daniela invested a total of $50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bo
LekaFEV [45]

Answer:

The amount invested in bonds = 35,000

The amount invested in CD = 15,000.

Step-by-step explanation:

The total amount that Daniela invested is $50,000, this means if we call the amount invested in bonds b, and the amount invested in CD d, then we have:

b+d=50,000 this says the total amount Daniela invested is $50,000.

And since the amount invested in bonds b is $5,000 more than twice the amount Daniela put into the CD, we have:

b=5,000+2d.

Thus, we have two equations and two unknowns b and d:

(1). b+d=50,000

(2). b=5,000+2d,

and we solve this system by substituting b from the second equation into the first:

b+d=50,000\\5,000+2d+d=50,000\\3d=45,000\\\\\boxed{d=15,000}

or, the amount invested in CD is $15,000.

With the value of d in hand, we now solve for b from equation(2):

b=5,000+2d\\b=5000+2(15,000)\\\boxed{b=35,000}

or, the amount invested in bonds is $35,000.

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Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)
vagabundo [1.1K]

Answer:

Considering the given equation y = log_{3}x\\

And the ordered pairs in the format (x, y)

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For (\frac{1}{3}, a_{0} )

x=\frac{1}{3}

y=a_{0}

y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1

So the ordered pair will be (\frac{1}{3}, -1 )

For (1, a_{1} )

x=1

y=a_{1}

y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0

So the ordered pair will be (1, 0 )

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x=3

y=a_{2}

y = log_{3}x\\y = log_{3}3\\y = 1

So the ordered pair will be (3, 1 )

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x=9

y=a_{3}

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So the ordered pair will be (9, 2 )

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y=a_{4}

y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3

So the ordered pair will be (27, 3 )

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So the ordered pair will be (81, 4 )

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Factor strings for 36
Marina CMI [18]
Factor strings for 36:

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4 x 9
2 x 2 x 3 x 3
2 x 2 x 9
4 x 3 x 3
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Answer:

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