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bezimeni [28]
4 years ago
7

Simplify the expression to a polynomial in standard form: (3x+1)(2x^2-5x+7)

Mathematics
1 answer:
Ilia_Sergeevich [38]4 years ago
5 0

Answer:

6x^3-13x^2+16x+7

Step-by-step explanation:

(3x+1)(2x^2-5x+7)

6x^3-15x^2+21x+2x^2-5x+7

6x^3-15x^2+2x^2+21x-5x+7

6x^3-13x^2+16x+7

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A certain company recently sold five-year $1000 bonds with an annual yield of 9.75%. After how much time could they be sold for
Brut [27]

Answer:

After 7 years and 5 months.

Step-by-step explanation:

Let x represent number of years.

We have been given that a certain company recently sold five-year $1000 bonds with an annual yield of 9.75%.

We can see that the value of bond is increasing exponentially, so we will use exponential growth formula to solve our given problem.

y=a\cdot (1+r)^x, where,

y = Final value,

a = Initial value,

r = Rate in decimal form,

x = Time

9.75\%=\frac{9.75}{100}=0.0975

Substituting given values:

y=1000\cdot (1+0.0975)^x

y=1000\cdot (1.0975)^x

Since we need the selling price to be twice the original price, so we will substitute y=2000 in above equation as:

2000=1000\cdot (1.0975)^x

\frac{2000}{1000}=\frac{1000\cdot (1.0975)^x}{1000}

2=1.0975^x

Switch sides:

1.0975^x=2

Take natural log of both sides:

\text{ln}(1.0975^x)=\text{ln}(2)

Applying rule \text{ln}(a^b)=b\cdot \text{ln}(a):

x\cdot \text{ln}(1.0975)=\text{ln}(2)

\frac{x\cdot \text{ln}(1.0975)}{\text{ln}(1.0975)}=\frac{\text{ln}(2)}{\text{ln}(1.0975)}

x=\frac{0.6931471805599453}{0.0930348659671894}

x=7.45040231265

x\approx 7.4504

Since x represents time in years, so we need to convert decimal part into months by multiplying .4504 by 12 as 1 year equals 12 months.

7 years and 12*0.4504023 months = 7 years 5.4 months = 7 years 5 months

Therefore, after 7 years and 5 months the company could sold the bonds for twice their original price.

8 0
3 years ago
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The Xerox tower has 30 stories
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drew has 32 toy cars. He trades 7 of those cars for 11 others toy cars. how many toy cars does drew have now.
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Drew has 36 cars now.
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What is a1 of the arithmetic sequence for which a3=126 and a64=3,725
Evgesh-ka [11]
D=an-am/n-m
d=a64-a3/64-3
d=3725-126/61=59
a3=a1+2d
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