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Tpy6a [65]
3 years ago
15

Are all imaginary numbers like terms

Mathematics
1 answer:
Sever21 [200]3 years ago
8 0
No there not hope this helps :)
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4 plus the product of 2 and 7<br>please help me
madreJ [45]

The answer is 18. Because 7*2=14+4=18

3 0
3 years ago
Alice's average mark for 4 Science exams is 68%. Can Alice score enough marks in a fifth exam so that her average mark increases
Keith_Richards [23]

Answer:

It is not possible for Alice to score an average of 75% because what she needs to score in the fifth exam exceeds the maximum possible score.

Step-by-step explanation:

She scored an average 68% in all 4 exams.

This means; (68% + 68% + 68% + 68%)/4

Now, a fifth exam is added and we want to know if her average can now be 75%.

Thus, let the score in the 5th exam be x and we now have;

(68% + 68% + 68% + 68% + x)/5 = 75%

Multiply both sides by 5 to get;

(68% + 68% + 68% + 68% + x) = 375%

272% + x = 375%

x = 375% - 272%

x = 103%

The maximum score is 100% and therefore it is impossible for her to score 103%. Therefore it is not possible for Alice to score an average of 75%

7 0
3 years ago
After selling some lemonade and cookies, Vivian and her brother Gil had 7-one dollar bills, 8 quarters and 6 dimes.They agreed t
vichka [17]

Answer:

They earned $9.60 they will each get $4.80. Give or take.

Step-by-step explanation:

9.60÷2=4.80

6 0
3 years ago
Help with 30 please. thanks.​
Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

\displaystyle  y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}

And we want to show that:

\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}

Instead of differentiating directly, we can first square both sides:

\displaystyle y^2 = 3e^{2x} -4x + 1

We can find the first derivative through implicit differentiation:

\displaystyle 2y \frac{dy}{dx}  = 6e^{2x} -4

Hence:

\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}

And we can find the second derivative by using the quotient rule:

\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}

Substitute:

\displaystyle y\left(\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\right) + \left(\frac{3e^{2x}-2}{y}\right)^2 =6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}- \left(3e^{2x} -2\right)^2}{y^2} + \frac{\left(3e^{2x}-2\right)^2}{y^2}= 6e^{2x}

Combine fractions:

\displaystyle \frac{\left(6y^2e^{2x}-\left(3e^{2x} - 2\right)^2\right) +\left(\left(3e^{2x}-2\right)^2\right)}{y^2} = 6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}}{y^2} = 6e^{2x}

Simplify:

6e^{2x} \stackrel{\checkmark}{=} 6e^{2x}

Q.E.D.

6 0
2 years ago
PLEASE HELP!!! CALCULUS ASSIGNMENT
Serga [27]

Answer:

(d)  \displaystyle 12x^3 - 15x^2 + 2

General Formulas and Concepts:

<u>Algebra I</u>

  • Functions
  • Function Notation

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]:                                                                \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = 3x^4 - 5x^3 + 2x - 1

<u>Step 2: Differentiate</u>

  1. Basic Power Rule:                                                                                            \displaystyle y' = 4(3x^{4 - 1}) - 3(5x^{3 - 1}) + 2x^{1 - 1} - 0
  2. Simplify:                                                                                                             \displaystyle y' = 4(3x^3) - 3(5x^2) + 2
  3. Multiply:                                                                                                             \displaystyle y' = 12x^3 - 15x^2 + 2

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

Unit: Derivatives

Book: College Calculus 10e

3 0
3 years ago
Read 2 more answers
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