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Diano4ka-milaya [45]
3 years ago
12

How to solve ...Please and thank you ​-5y^2=105

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
5 0

- 5 {y}^{2}  = 105 \\\Leftrightarrow  {y}^{2}  =  - 21 \\ \Leftrightarrow y = i \sqrt{21}

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Explain the error a student is asked to find when the value of an investment of $5200 is an account that earns 4.2% annual inter
zloy xaker [14]

We are given the following information

Investment = $5200

Annual interest rate = 4.2% = 0.042

Final amount = $16,500

Number of years = 27.7 years

Number of compoudings = quartely = 4

The student uses the following model

V(t)=5200(1.014)^{3t}

The general formula for compound interest is given by

V(t)=P(1+\frac{r}{n})^{nt}

As you can see, the number of compoundings is incorrect (3 vs 4)

The interest rate is also incorrect.

Let us substitute the given values into the above formula

\begin{gathered} V(t)=P(1+\frac{r}{n})^{nt} \\ V(t)=5200(1+\frac{0.042}{4})^{4\cdot27.7} \\ V(t)=5200(1+0.0105)^{110.8} \\ V(t)=5200(1.0105)^{110.8} \\ V(t)=\$16543.5^{} \end{gathered}

Therefore, the final amount is approximately $16,543.5

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sleet_krkn [62]

Answer:

Option A

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Answer:

Step-by-step explanation:

Hello, first, let's use the product rule.

Derivative of uv is u'v + u v', so it gives:

f(x)=(x^3-2x+1)(x-3)=u(x) \cdot v(x)\\\\f'(x)=u'(x)v(x)+u(x)v'(x)\\\\ \text{ **** } u(x)=x^3-2x+1 \ \ \ so \ \ \ u'(x)=3x^2-2\\\\\text{ **** } v(x)=x-3 \ \ \ so \ \ \ v'(x)=1\\\\f'(x)=(3x^2-2)(x-3)+(x^3-2x+1)(1)\\\\f'(x)=3x^3-9x^2-2x+6 + x^3-2x+1\\\\\boxed{f'(x)=4x^3-9x^2-4x+7}

Now, we distribute the expression of f(x) and find the derivative afterwards.

f(x)=(x^3-2x+1)(x-3)\\\\=x^4-2x^2+x-3x^3+6x-4\\\\=x^4-3x^3-2x^2+7x-4 \ \ \ so\\ \\\boxed{f'(x)=4x^3-9x^2-4x+7}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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That means she need to buy 2 packs of 100.
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