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aleksklad [387]

We know that we have to make a down payment of $1500 and finance the rest of $20000 at a 1.9% interest rate, making equal monthly payments for 5 years. Our first step to solve this problem would be to convert 5 years into months.

1 year = 12 months

12 * 5 = 60 months

Therefore, in 5 years there are 60 months.

Now lets solve this problem step by step.

Subtract the down payment from $20,000

$20000-$1500=$18500

Multiply the remaining number by the interest rate.

$18500 *1.9 = $35150

Divide 35150 by number of months in 5 years (60)

$35150 / 60 = $585

Therefore, you have to pay $585 per month.

6 0

2 years ago

Lucy spends 4 hours a week babysitting. Her sister Lily spends seven eighths as much time babysitting. Does Lily babysit for mor

olganol [36]

Answer:

Lily babysit for less than 4 hours.

Step-by-step explanation:

Given:

Lucy spends in a week babysitting for 4 hours.

Lily spends seven-eighths of 4 hours.

Now, to find whether Lily babysit for more than 4 hours or less than that.

Number of hours Lucy babysit = 4 hours.

So, to get the hours Lily babysit:

$\frac{7}{8}\ of\ 4\ hours.$

$=\frac{7}{8}\times 4\\\\=\frac{28}{8} \\\\$

On simplifying we get:

$=\frac{7}{2} \\\\=3\frac{1}{2}\ hours.$

Thus, Lily spends $3\frac{1}{2}\ hours$ for babysitting which is less than 4 hours.

Therefore, Lily babysit for less than 4 hours.

4 0

3 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$