All categories

3 years ago

Will you please help me

Mathematics

2 answers:

UkoKoshka [18]3 years ago

6 0

Answer: B

Step-by-step explanation:

In the image below answer b is in color green and the rest of the answers are clear straight lines making them linear.

horsena [70]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Given the rectangle abcd shown below has a total area of 72. E is in the midpoint of bc and f is the midpoint of dc. What is the

scoray [572]

Refer to the attached image.

Given the rectangle ABCD of length 'l' and height 'h'.

Therefore, CD=AB = 'l' and BC = AD = 'h'

We have to determine the area of triangle AEF.

Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

Area of triangle ADF = $\frac{1}{2}bh$

= $\frac{1}{2}(DF \times AD)$

= $\frac{1}{2}(\frac{l}{2} \times h)$

$=\frac{lh}{4}$

Area of triangle ECF = $\frac{1}{2}bh$

= $\frac{1}{2}(CF \times CE)$

= $\frac{1}{2}(\frac{l}{2} \times \frac{h}{2})$

$=\frac{lh}{8}$

Area of triangle ABE = $\frac{1}{2}bh$

= $\frac{1}{2}(AB \times BE)$

= $\frac{1}{2}(l \times \frac{h}{2})$

$=\frac{lh}{4}$

Now, area of triangle AEF =

Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

= $72 - (\frac{lh}{4} + \frac{lh}{8} + \frac{lh}{4})$

= $72 - (\frac{2lh+lh+2lh}{8})$

= $72 - (\frac{5lh}{8})$

= $72 - (\frac{5 \times 72}{8})$

$=\frac{72 \times 8 - (5 \times 72)}{8}$

= 27 units

Therefore, the area of triangle AEF is 27 units.

8 0

3 years ago

5. On a business trip Bob traveled 245.22 miles. If gasoline was $2.87 per gallon and the car averaged 25.3 miles per gallon, de

Alecsey [184]

The total cost of the trip was $27.81

Step-by-step explanation:

Given,

Distance traveled = 245.22 miles

Distance traveled per gallon = 25.3 miles

Gallons used for 245.22 miles = $\frac{Total\ distance}{Distance\ per\ gallon}$

Gallons used for 245.22 miles = $\frac{245.22}{25.3} = 9.69\ gallons$

9.69 gallons were used for 245.22 miles.

Cost per gallon = $2.87

Cost of 9.69 gallons = 2.87*9.69 = $27.81

The total cost of the trip was $27.81

Keywords: division, multiplication

Learn more about division at:

#LearnwithBrainly

4 0

3 years ago

Prove the following integration formula:

7nadin3 [17]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Distributive Property
Equality Properties

Algebra I

Combining Like Terms
Factoring

Calculus

Derivative 1: $\frac{d}{dx} [e^u]=u'e^u$
Integration Constant C
Integral 1: $\int {e^x} \, dx = e^x + C$
Integral 2: $\int {sin(x)} \, dx = -cos(x) + C$
Integral 3: $\int {cos(x)} \, dx = sin(x) + C$
Integral Rule 1: $\int {cf(x)} \, dx = c \int {f(x)} \, dx$
Integration by Parts: $\int {u} \, dv = uv - \int {v} \, du$
[IBP] LIPET: Logs, Inverses, Polynomials, Exponents, Trig

Step-by-step Explanation:

Step 1: Define Integral

$\int {e^{au}sin(bu)} \, du$

Step 2: Identify Variables Pt. 1

Using LIPET, we determine the variables for IBP.

Use Int Rules 2 + 3.

$u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{-cos(bu)}{b}$

Step 3: Integrate Pt. 1

Integrate [IBP]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} - \int ({ae^{au} \cdot \frac{-cos(bu)}{b} }) \, du$
Integrate [Int Rule 1]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} \int ({e^{au}cos(bu)}) \, du$

Step 4: Identify Variables Pt. 2

Using LIPET, we determine the variables for the 2nd IBP.

Use Int Rules 2 + 3.

$u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b}$

Step 5: Integrate Pt. 2

Integrate [IBP]: $\int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \int ({ae^{au} \cdot \frac{sin(bu)}{b} }) \, du$
Integrate [Int Rule 1]: $\int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du$

Step 6: Integrate Pt. 3

Integrate [Alg - Back substitute]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} [\frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du]$
[Integral - Alg] Distribute Brackets: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2} - \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du$
[Integral - Alg] Isolate Original Terms: $\int {e^{au}sin(bu)} \, du + \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du= \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}$
[Integral - Alg] Rewrite: $(\frac{a^2}{b^2} +1)\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}$
[Integral - Alg] Isolate Original: $\int {e^{au}sin(bu)} \, du = \frac{\frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +1}$
[Integral - Alg] Rewrite Fraction: $\int {e^{au}sin(bu)} \, du = \frac{\frac{-be^{au}cos(bu)}{b^2} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +\frac{b^2}{b^2} }$
[Integral - Alg] Combine Like Terms: $\int {e^{au}sin(bu)} \, du = \frac{\frac{ae^{au}sin(bu)-be^{au}cos(bu)}{b^2} }{\frac{a^2+b^2}{b^2} }$
[Integral - Alg] Divide: $\int {e^{au}sin(bu)} \, du = \frac{ae^{au}sin(bu) - be^{au}cos(bu)}{b^2} \cdot \frac{b^2}{a^2 + b^2}$
[Integral - Alg] Multiply: $\int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)]$
[Integral - Alg] Factor: $\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)]$
[Integral] Integration Constant: $\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C$