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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!!

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

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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>

Evaluate Exponents: $\frac{-5 \pm \sqrt{25-4(3)(1)} }{2(3)}$
Evaluate Multiplication: $\frac{-5 \pm \sqrt{25-12} }{6}$
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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3 years ago

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Which is the best example of a short-term savings goal?

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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 $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

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

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

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

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$

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Factor strings for 36

Marina CMI [18]

Factor strings for 36:

6 x 6
2 x 18
3 x 12
4 x 9
2 x 2 x 3 x 3
2 x 2 x 9
4 x 3 x 3
6 x 2 x 3

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Answer:

c = -20

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