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3 years ago

Let f( x)=2x^2+7x and g(x)=-8x^2-3x+5. What is g(x)+f(x)?

Mathematics

1 answer:

Natalka [10]3 years ago

3 0

Answer:

-6x^2+4x+5

Step-by-step explanation:

g(x)+f(x)=(-8x^2-3x+5)+(2x^2+7x)

=-8x^2-3x+5+2x^2+7x

=-8x^2+2x^2-3x+7x+5

=-6x^2+4x+5

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Can someone please help me

Irina-Kira [14]

Using exponentials, the expression can be presented as the following:

= 7^(1/3) * 7^(1/2) / 7^(1/6)= 7^(1/3 + 1/2 - 1/6)= 7^(2/6 + 3/6 - 1/6)= 7^(2/3)

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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: $\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C$

And we have proved the integration formula!

6 0

3 years ago

Read 2 more answers

Write the equation of the line perpendicular to

r-ruslan [8.4K]

Solving for y, we add 5y to both sides and subtract 4, getting 9x-4=5y. Dividing both sides by 5, we get 9x/5-4/5=y. Since the slope is 9/5 (since 9/5*x=9x/5), we multiply it by -1 and find the reciprocal of it to get -5/9 as the perpendicular slope, so -5x/9+b=y. Plugging 1 in for x and -6 in for y, we get -5*1/9+b=-6 and by adding 5/9 to both sides we get -5-4/9=b , and since in y=mx+b y and x are variables, we end up with y=-5x/9+(-5-4/9) for slope intercept form.

To get it into standard form, we need it in ay+cx=b with a, b, and c being constants. Adding 5x/9 to both sides, we end up with y+5x/9=(-5-4/9) for standard form

5 0

3 years ago

y- 8x-2 and y Rajib wrote the equations5. What can Rajib conclude about the solution to this system ofequations?o The point (2,1

ale4655 [162]

Its just a matter of subbing in ur answer choices to see which ones are correct...but remember, for it to be a solution, it has to satisfy BOTH equations.

(-1/4,-4)
y = 8x - 2
-4 = 8(-1/4) - 2
-4 = - 8/4 - 2
-4 = -2-2
-4 = -4 (correct)

(-1/4,-4)
y = -4x - 5
-4 = -4(-1/4) - 5
-4 = 1 - 4
-4 = -4 (correct)

Therefore, ur solution is (-1/4,-4)

8 0

3 years ago

Read 2 more answers

Solve to find x and y in the diagram. HELP ASAP!!

Lapatulllka [165]

X=12, y=7.5

Here is an explanation if you want to know how to do it:

First, look at the diagram and think about what you know so far. Since the lines are parallel, 5x+4y and 12y must be equal. We also know that there is a right angle, so since both angles are equal, 12y=90 and 5x+4y=90.

Solve for y by dividing 90 by 12 to get 7.5, so y=7.5. Sub in this value to the equation on top and you should get 5x+4(7.5)=90 4(7.5)=30, so subtract that from 90 to get 60 and divide 60 by 5 to get 12, so x=12.

5 0

3 years ago