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2 years ago

Here's a brain teaser. (I already know the answer) 1=5 2=10 3=15 4=20 5=?

Mathematics

2 answers:

prisoha [69]2 years ago

8 0

Answer:

5 = 1

5 = 25

Step-by-step explanation:

I've seen two answers to this.

1. Since we are told 1 = 5, then 5 = 1.

One answer is 1 = 5

2. Since each number on the right is 5 times the number on the left, then 25 corresponds to 5, and 5 = 25.

aivan3 [116]2 years ago

4 0

Answer: 5=1 (The correct answer) 5=25 (The answer that makes sense but is not correct).

Step-by-step explanation:

Explanation for answer 1: It say's that 1 = 5.

Explanation for answer 2: If we follow the pattern of adding 5 each time we go up a number then 5 would equal 25.

You might be interested in

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

See Photo for information on how to answer

stiks02 [169]

Answer:

It is a Geometric sequence

Common ratio = 1/5

Step-by-step explanation:

Whole chain: 25, 5, 1, 1/5, 1/25, 1/125, 1/625, 1/3125

Hope this helped

6 0

3 years ago

Find the inverse please and thank you

Amanda [17]

The inverse $f^{-1}(x)$ is such that

$f\left(f^{-1}(x)\right) = x$

We have

$f\left(f^{-1}(x)\right) = -9\sqrt{f^{-1}(x) - 8} + 5 = x$

Solve for the inverse.

$-9\sqrt{f^{-1}(x)-8} + 5 = x \\\\ -9\sqrt{f^{-1}(x)-8} = x-5 \\\\ \sqrt{f^{-1}(x)-8} = -\dfrac{x-5}9 \\\\ \left(\sqrt{f^{-1}(x)-8}\right)^2 = \left(-\dfrac{x-5}9\right)^2 \\\\ f^{-1}(x) - 8 = \dfrac{(x-5)^2}{81} \\\\ \boxed{f^{-1}(x) = 8 + \dfrac{(x-5)^2}{81}}$

5 0

2 years ago

A triangle has side lengths of (7x-4)(7x−4) centimeters, (x+3)(x+3) centimeters, and (3y+2)(3y+2) centimeters. Which expression

Papessa [141]

Answer:

This assumes that the question was meant to read:

"A triangle has side lengths of (7x−4) centimeters, (x+3) centimeters, and (3x+2) centimeters." And not "A triangle has side lengths of (7x-4)(7x−4) centimeters, (x+3)(x+3) centimeters, and (3y+2)(3y+2) centimeters."

If the revision is correct, the perimeter is 11x + 1 cm

=============================================

If the question was correctly written, then answer is:

50x^2 - 50x + 9y^2 + 12 y + 29

===============

Step-by-step explanation:

(7x-4) cm

+ (x+3) cm

+ (3x+2) cm

11x + 1 cm

If the question was correctly written as is, then:

(7x-4)(7x−4) = 49x^2 - 56x + 16

+(x+3)(x+3) = + x^2 + 6x + 9

+(3y+2)(3y+2) = +9y^2 + 12 y + 4

50x^2 - 50x + 9y^2 + 12 y + 29 cm

8 0

2 years ago

Who play basketball?

babymother [125]

Answer:

LeBron James

Step-by-step explanation:

3 0

3 years ago