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Makovka662 [10]
1 year ago
10

Evaluate f (x) = 3 (2)x for x=-1

Mathematics
1 answer:
kap26 [50]1 year ago
8 0

The value of the function at x = -1 is 5

<h3>What is  function?</h3>

Function can be defined as an expression, rule, or law that defines a relationship between one variable known as the independent variable and another called the dependent variable.

From the information given, we have that;

f (x) = 3 (2)x  with  x=-1

Now, let's substitute the value of x as - 1 in the function given;

f (x) = 3 (2)x

f (x) = 3 (2) -1

Expand the bracket;

f (x) = 6 - 1

f(x) = 5

Thus, the value of the function at x = -1 is 5

Learn more about functions here:

brainly.com/question/6561461

#SPJ1

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You make $5000 per morth. If you spend 50% on necessitles, 30% on stuff you want and 20% of it you end up saving, how much money
liubo4ka [24]

Answer:

$1000 per month

Step-by-step explanation:

5000 dollars is 100 percent so we put 5000 over 100

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then since we need to find 20 percent of something, we put x over 20

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Then we cross multiply

<u>5000</u>      times       <u>x   </u>            

100                        20

so 100x = 100000

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Good luck, and hope I did this correctly

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Point A(7,3) is translated to A' (16.-9). Which rule describes the translation? IO (X)(-9.7-12) O (x,y)-&gt; (x-9.y+12) 0 (x,y)
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Answer:

(x+9,y-12)

Step-by-step explanation:

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3 0
3 years ago
Improper fraction for 3 1/14
TEA [102]

Answer: 43/14

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So 3 times 14=42

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On the 1st January 2014 Carol invested some money in a bank account.
Ghella [55]

Answer:

\large \boxed{\text{\pounds 23 360.00}}

Step-by-step explanation:

The formula for the accrued amount from compound interest is

A = P \left(1 + \dfrac{r}{n}\right)^{nt}

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:  

r = 0.025

\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}

3. Summary

1 Jan 2014      P = £23 360.00

1 Jan 2015     A =    23 944.00

     Withdrawal = <u>    -1  000.00 </u>

                     P =     22 944.00

1 Jan 2016    A =    £23 517.60

5 0
3 years ago
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