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morpeh [17]
3 years ago
8

If you bought a stock last year for a price of $67, and it has risen 12% since then, how much is the stock worth now, to the nea

rest cent?
Mathematics
1 answer:
lisabon 2012 [21]3 years ago
4 0

Answer:$75.04

Step-by-step explanation:

last year price=$67

12% of 67

12/100 x 67

(12 x 67) ➗ 100

804 ➗ 100=8.04

Since it has risen by 12%

There new price is 67+8.04=75.04

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The number of camera phones shipped globally can be modeled by the function y = 1.28e 1.31x where x is the number of years since
OlgaM077 [116]

Answer: 895 phones

Step-by-step explanation:

Given that :

The function : y = 1.28e^1.31x ; is used to model the number of camera phones shipped since 1997

y = number of camera phones ; x = number of years since 1997

The number of camera phones shipped in the year 2002 can be obtained thus ;

x = 2002 - 1997 = 5 years

y = 1.28e^(1.31 * 5)

y = 1.28e^6.55

y = 1.28(699.24417)

y = 895.03254

y = 895 phones

3 0
3 years ago
Write a rule for the function whose graph can be obtained from the given parent function by performing the given transformations
kati45 [8]

ANSWER

g(x) =   {(x + 5)}^{3}  + 4

EXPLANATION

Given the parent function:

f(x) =  {x}^{3}

If we shift the graph of f(x), 5 units to the left, then the new rule is

x \to \:  {(x + 5)}^{3}

If we again, shift this graph upward by 4 units, then the completely transformed function will have the rule,

x \to \:  {(x + 5)}^{3}  + 4

Or

g(x) =   {(x + 5)}^{3}  + 4

7 0
3 years ago
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Answer:

y=3x-7

step by step explaination:

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6 0
4 years ago
The cost of a customer smartphone bill can be modeled by the equation c=10g+50
Vika [28.1K]

Answer:

c=10(g+5)

Step-by-step explanation:

Given that,

The cost of customer smartphone bill is given by the equation as follows :

c=10g+50

Where

g is the number of gigabits of data the customer per month

We need to factor this cost polynomial

The common between 10 and 50 is 10

Taking 10 common. So,

c=10(g+5)

Hence, the required factor is equal to c=10(g+5).

6 0
3 years ago
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