Are all imaginary numbers like terms

4 plus the product of 2 and 7 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

3 years ago

PLEASE HELP!!! CALCULUS ASSIGNMENT

Serga [27]

Answer:

(d) $\displaystyle 12x^3 - 15x^2 + 2$

General Formulas and Concepts:

Algebra I

Functions
Function Notation

Calculus

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:

Step 1: Define

Identify

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

Step 2: Differentiate

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

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