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3 years ago

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A car mechanic borrow $7000 for 12 months at 13% interest rate the monthly payments are $817.43 what is the total amount includi

ng interest (repayment), that the mechanic paid for loan

1 answer:

hoa [83]3 years ago

3 0

It is 90,0000000000000000000000

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you and your friend start biking to opposite directions from the same point. You travel 108 feet every 8 seconds. Your friend tr

juin [17]

Answer:

To find your distance apart, you can convert 15 minutes to seconds because that is what the rate is given in.

15 mins x 60 seconds = 900 seconds.

Step-by-step explanation:

In 900 seconds, there are 900/8=112.5 groups of 8 seconds. This means there are 112.5 groups of 108 feet.

112.5 x 108 = 12150 feet

12150 feet/5280 feet is approximately 2.3 miles.

For your friend you take 900/6 = 150 groups of 6 seconds.

150 x 63 = 9540 feet

9540 feet/5280 feet = 1.79 miles

1.79 miles + 2.3 miles = 2.09 miles

You and your friend are about 2.09 miles apart after 15 minutes.

B). If you go 2.3 miles every 15 minutes, that means you travel about 0.77 miles every 5 minutes. If you travel 20 more minutes (40-15-5) that would be 0.77x 4=3.07 miles more

3.07 + 2.3 = 5.37 miles after 40 minutes (you).

Your friend travels 1.79 in 15 minutes, so 1.79/3 = 0.6 miles every 5 minutes

0.6 x 5 (5 x 5 = 25 minutes)= 3 miles

1.79 + 3 miles= 4.79 miles (friend)

4.79 + 5.37 = 10.16 miles

You and your friend are about 10.16 miles apart after 40 minutes.

5 0

3 years ago

Express 10cm as percentage of 50cm

tino4ka555 [31]

I think that 10 is 20% of 50

4 0

2 years ago

Read 2 more answers

If there are 14 dogs and 23 cats in the shelter. Which ratio DOES NOT accurately compare the

Leona [35]

Answer: B. 23 to 14

Since it asks for the ratio of dogs to cats the 14 has to go first. The other options are all different ways to write the same ratio. Hope this helps!

6 0

3 years ago

Find the area of the region bounded by the graphs of y = x, y = −x + 4, and y = 0.

Marrrta [24]

The line y = x and y = -x + 4 intersect when at the point (2, 2).

Expresing y = -x + 4 in terms of x, we have x = 4 - y.

Thus, the area of the region bounded by the <span>graphs of y = x, y = −x + 4, and y = 0 is given by

$\int\limits^2_0 {(y-(4-y))} \, dy = \int\limits^2_0 {(y-4+y)} \, dy \\ \\ = \int\limits^2_0 {(2y-4)} \, dy= \left[y^2-4y\right]_0^2 =|(2)^2-4(2)| \\ \\ =|4-8|=|-4|=4$

Therefore, the area bounded by the lines is 4 square units.
</span>

4 0

3 years ago

Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)

sergiy2304 [10]

Answer:

$\displaystyle A = \frac{20\sqrt{15}}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

<u>Algebra I</u>

Terms/Coefficients
Graphing
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Area - Integrals

U-Substitution

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

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

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

<u>Step 2: Find Bounds of Integration</u>

<em>Solve each equation for the x-value for our bounds of integration.</em>

F

Set <em>y</em> = 0: 0 = √(15 - x)
[Equality Property] Square both sides: 0 = 15 - x
[Subtraction Property of Equality] Isolate <em>x</em> term: -x = -15
[Division Property of Equality] Isolate <em>x</em>: x = 15

G

Set y = 0: 0 = √(15 - 3x)
[Equality Property] Square both sides: 0 = 15 - 3x
[Subtraction Property of Equality] Isolate <em>x</em> term: -3x = -15
[Division Property of Equality] Isolate <em>x</em>: x = 5

This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.

We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).

<u>Step 3: Find Area of Region</u>

<em>Integration Part 1</em>

Rewrite Area of Region Formula [Integration Property - Subtraction]: $\displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx$
[Integral] Substitute in variables and limits [Area of Region Formula]: $\displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx$
[Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

<u>Step 4: Identify Variables</u>

<em>Set variables for u-substitution for both integrals.</em>

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

<u>Step 5: Find Area of Region</u>

<em>Integration Part 2</em>

[Area] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx$
[Area] U-Substitution: $\displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz$
[Area] Reverse Power Rule: $\displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})$
[Area] Multiply: $\displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}$
[Area] Add: $\displaystyle A = \frac{20\sqrt{15}}{3}$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

3 0

3 years ago