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Shtirlitz [24]
3 years ago
6

Q.16 please With steps

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0

The equations

2x^2+9x-5=0

and

2\left(t-\dfrac12\right)^2+9\left(t-\dfrac12\right)-5=0

are really the same, we're just setting x=t-\dfrac12, or t=x+\dfrac12.

2x^2+9x-5=(2x-1)(x+5)=0\implies x=\dfrac12,x=-5

So we get

t=\dfrac12+\dfrac12=1,t=-5+\dfrac12=-\dfrac92

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Natasha_Volkova [10]

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4 years ago
Read 2 more answers
For p = $3000, r = 3.5%,and t = 5 years, Find the balance in an account When interest is compounded (a) quarterly, (b) monthly,a
MAVERICK [17]

Answer:

a) $3,571.02

b) $3,572.9

c) $3,573.74

Step-by-step explanation:

Data provided in the question:

p = $3000,

r = 3.5%,

t = 5 years

a) quarterly

number of periods in a year, n = 4

Interest rate per period = 3.5% ÷ 4 = 0.875%

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

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on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 4 } \right)^{\Large{ 4 \cdot 5 }}

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A = $3,571.02

b) monthly

number of periods in a year, n = 12

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 12 } \right)^{\Large{ 12 \cdot 5 }}

A = 3000 × [/tex]\cdot { 1.002917} ^ { 60 }[/tex]

A = 3000 × 1.190967

A = $3,572.9

c) continuously

A = pe^{r\times t}

on substituting the respective values, we get

A = 3,000 × e^{0.035\times 5}

or

A = 3,000 × e^{0.175}

or

A = 3,000 × 1.1912

or

A = $3,573.74

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3 years ago
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