Is (n - 1)^2 odd or even

Answer:

A

Step-by-step explanation:

We are given the equation:  

[tex]x^2+xy-3y=3[/tex]

And we want to find dy/dx at the point (2, 1).

Find dy/dx. We can take the derivative of both sides with respect to x:

[tex]\displaystyle \frac{d}{dx}[x^2+xy-3y]=\frac{d}{dx}[3][/tex]

Implicitly differentiate:

[tex]\displaystyle 2x+y+x\frac{dy}{dx}-3\frac{dy}{dx}=0[/tex]

Solve for dy/dx:

[tex]\displaystyle \begin{aligned} x \frac{dy}{dx} - 3\frac{dy}{dx} & = -2x - y \\ \\ \frac{dy}{dx}(x - 3) & = -2x - y \\ \\ \frac{dy}{dx} & = \frac{-2x-y}{x-3} \\ \\ & = \frac{2x+y}{3-x} \end{aligned}[/tex]

To find dy/dx at (2, 1), evaluate dy/dx for x = 2 and y = 1. Hence:

[tex]\displaystyle \frac{dy}{dx}\Big|_{(2, 1)}=\frac{2(2)+1}{3 - (2)}=\frac{5}{1}=5[/tex]

Hence, our answer is A.

We want to find the differential dy/dx for the given expression at the given point. We will see that the correct option is A.

We have:

x^2 + xy - 3y = 3

First, we need to isolate the variable y in one side of the equation, so we get:

xy - 3y = 3 - x^2

y*(x - 3) = 3 - x^2

y = (x - x^2)/(x - 3)

Now we can differentiate this, remember that if:

f(x) = g(x)*h(x)

then:

f'(x) = g(x)*h'(x) + g'(x)*h'(x)

Using that rule.

dy/dx = (1 - 2x)/(x - 3) - (x - x^2)/(x - 3)^2

Now we need to evaluate this in x = 2 (the x-value of the point) so we get:

(1 - 2*2)/(2 - 3) - (2 - 2^2)/(2 - 3)^2

= (-3)/(-1) - (-2)/(1) = 5

So the correct option is A.

If you want to learn more about differentiation, you can read:

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Answer:

5/4, 11/12, 1/4, 5/12, 4/10        0.91, 0.89, 0.51, 0.5, 0.04

Step-by-step explanation:

Im pretty sure these are correct

Answer:

C

Step-by-step explanation:

Answer:

it turns out google has the answer

Step-by-step explanation:

can i get brainlyest

Answer:

I think the answer is

x  = 6, -2/3

Step-by-step explanation:

I hope this helps please let me know if it is incorrect

Answer:

88.9%

Step-by-step explanation:

Answer:

6 units

Step-by-step explanation:

BC = BD/2 = 8

PB² + 8² = 10²

PB² = 100 - 64 = 36 ⇒ PB = 6 units

The length of PB = 6 units

Here, PB is the perpendicular bisector of AC and QC is the perpendicular bisector of BD.

AB = BC = CD

Given that BD = 16 and PC = 10

What is Pythagoras theorem?

Pythagoras theorem is well known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square of the hypotenuse.

Here, BD = 16 and AB = BC = CD

⇒ BC = BD/2 = 16/2 = 8

And, PC = 10

Now using Pythagoras theorem:

[tex]PC^{2} =PB^{2} + BC^{2}[/tex]

[tex]10^{2} =PB^{2} + 8^{2}[/tex]

[tex]PB^{2} + 64 = 100\\PB^{2} = 100 - 64\\PB^{2} = 36\\PB = \sqrt{36}\\PB = 6[/tex]

Hence the length of PB = 6 Units

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Answer: Approximately 51 years.

Step-by-step explanation: A population grows by changing exponentially over time, which can be mathematically demonstrated as:

[tex]P=P_{0}e^{rt}[/tex]

where

P is the population after time

P₀ is initial population, when time = 0

r is a percentage rate of growth

t is time passed

In this case, we have to calculate the amount of time it has passed for a population to double, so [tex]P=2P_{0}[/tex]:

[tex]2P_{0}=P_{0}e^{0.0135t}[/tex]

[tex]e^{0.0135t}=2[/tex]

[tex]ln(e^{0.0135t})=ln2[/tex]

Using Logarithm Rule [tex]ln(e^{k})=k[/tex]:

[tex]0.0135t=0.693[/tex]

t = 51.34

For a population with rate of 1.35%, it will take approximately 51 years to double.

Answer:

he should buy 3 gallons

Step-by-step explanation:

There are 16 cups in a gallon, and if each person will eat 2 cups, and there are 24 ppl, multiply 24 by 2 to get the number of cups.

24 x 2 = 48.

Then, divide that by 16 (the number of cups in a gallon) to t how many gallons.

48 ÷ 16 = 3

so, he needs 3 gallons

hope this helped!  

Answer:

3 hours

Step-by-step explanation:

20 ÷ 5 = 4 mph

12 ÷ 4 = 3 hours

hope this explanation makes sense! :)

Sara took 3 hours to hike 12 miles if, on the first day of camp, Sara hiked 20 miles in 5 hours and the rate is 4 miles per hour.

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

It is given that:

On the first day of camp, Sara hiked 20 miles in 5 hours.

The rate = 20/5 = 4 miles per hour

Let x be the time taken by Sara to hike 12 miles

As its rate of speed of Sara is 4 miles per hour

x = 12/4

The arithmetic operation can be defined as the operation in which we do the addition of numbers, subtraction, multiplication, and division.

x = 3 hours

Thus, Sara took 3 hours to hike 12 miles if, on the first day of camp, Sara hiked 20 miles in 5 hours and the rate is 4 miles per hour.

Learn more about the fraction here:

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The answer fam is........x^3 - 5x^2 + 2x

Answer:

i got D

Step-by-step explanation:

sin 67 = sin 52

21 BC

BC=17.98

BC=18

Answer: $8.4

Step-by-step explanation:

To find 30% of 12 first, we move the decimal point to the left two times. Now we have .3 next multiply it by twelve. (3.6) This number is 30% of 12. Lastly, subtract 3.6 from 12 to get your answer.

Answer:

4:14  2:7    8:28

Step-by-step explanation:

Answer:

its 61.4x+15

Step-by-step explanation:

Step-by-step explanation:

let age of Byron be b yr old and Cindy be c yr old.

c = b + 2 --(1)

b - 8 + c - 8 = 14

b + c = 30 --(2)

sub (1) into (2):

2b = 28

b = 14

therefore Byron is 14 yr old now.

Topic: simultaneous eqns

If you like to venture further, do check out my insta (learntionary) where I regularly post useful math tips! Thank you!

Answer:

10

Step-by-step explanation:

Byron is 2 yrs younger than Cindy. 8 years ago the sum of their ages was 14.

Byron is 2 years at that time. Definitely, Cindy is 12 years old so the sum of their ages was 14. Let us just add 8 years to Byron's age at that time so 2 + 8 = 10

P.S: This just based on my understanding of the problem.

Answer:

B. 4 mL

Step-by-step explanation:

Answer:

5418.54

Step-by-step explanation:

because yes

Answer:31

Step-by-step explanation:

Answer:

b) 31 %

Step-by-step explanation:

The answer is 0.7%

Hope this helped!

Answer:

0.6%

Step-by-step explanation:

Percentage = 24/4000 *100

=0.6%

Answer:

375/1,000

Step-by-step explanation:

3/8=0.375

375/1,000 =0.375

Answer:

a. What does the Greene family pay as an annual premium? $450.60

b. How much would it cost him to purchase insurance in the amount of 100/300/25? $480.70

c. What is the difference in cost? $30.10

(っ◔◡◔)っ ♥here you go♥

Answer:

[tex]f(x) = -14x + 14[/tex]

Step-by-step explanation:

Given

[tex]Initial\ Value = 14[/tex]

[tex]Rate = -14[/tex]

Required:

Write as a function

Let x be the quantity that changes:

A function is of the form:

[tex]f(x) = mx + b[/tex]

Where

m = slope or rate

b = y intercept or initial value.

So, we can say that:

[tex]m = -14[/tex]

[tex]b = 14[/tex]

Substitute values for b and m in [tex]f(x) = mx + b[/tex]

[tex]f(x) = -14x + 14[/tex]

Answer:

10/9

Step-by-step explanation:

1/3+7/9=10/9

Step-by-step explanation:

= 1 + 7

3 9

= 9+21

27

= 28

27 ans

Answer:

See explanation

Step-by-step explanation:

Solving (a):

[tex]a_n = 6a_{n-1}[/tex] where [tex]a_0 = 2[/tex]

n = 1

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_1 = 6a_{1-1}[/tex]

[tex]a_1 = 6a_{0}[/tex]

Substitute 2 for [tex]a_0[/tex]

[tex]a_1= 6 * 2[/tex]

[tex]a_1= 12[/tex]

n = 2

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_2 = 6a_{2-1}[/tex]

[tex]a_2 = 6a_{1}[/tex]

Substitute 12 for [tex]a_1[/tex]

[tex]a_2= 6 * 12[/tex]

[tex]a_2= 72[/tex]

n = 3

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_3 = 6a_{3-1}[/tex]

[tex]a_3 = 6a_2[/tex]

Substitute 72 for [tex]a_2[/tex]

[tex]a_3= 6 * 72[/tex]

[tex]a_3= 432[/tex]

n = 4

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_4 = 6a_{4-1}[/tex]

[tex]a_4 = 6a_{3}[/tex]

Substitute 432 for [tex]a_3[/tex]

[tex]a_4 = 6 * 432[/tex]

[tex]a_4 = 2592[/tex]

n = 5

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_5 = 6a_{5-1}[/tex]

[tex]a_5 = 6a_4[/tex]

Substitute 2592 for [tex]a_4[/tex]

[tex]a_5 = 6 * 2592[/tex]

[tex]a_5 = 15552[/tex]

n = 6

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_6 = 6a_{6-1}[/tex]

[tex]a_6 = 6a_{5}[/tex]

Substitute 15552 for [tex]a_5[/tex]

[tex]a_6 = 6 * 15552[/tex]

[tex]a_6 = 93312[/tex]

[tex]a_1= 12[/tex]   [tex]a_2= 72[/tex]   [tex]a_3= 432[/tex]   [tex]a_4 = 2592[/tex]   [tex]a_5 = 15552[/tex]   [tex]a_6 = 93312[/tex]

Solving (b):

[tex]a_n = a^2_{n - 1[/tex] where [tex]a_1 = 2[/tex]

We have

[tex]a_1 = 2[/tex] which serves as the first term

n =2

[tex]a_n = a^2_{n - 1[/tex]

[tex]a_2 = a^2_{2-1}[/tex]

[tex]a_2 = a^2_{1}[/tex]

Substitute 2 for [tex]a_1[/tex]

[tex]a_2 = 2^2[/tex]

[tex]a_2 = 4[/tex]

n = 3

[tex]a_3 = a^2_{3-1}[/tex]

[tex]a_3 = a^2_{2}[/tex]

[tex]a_3 = 4^2[/tex]

[tex]a_3 = 16[/tex]

n = 4

[tex]a_4 = a^2_{4-1}[/tex]

[tex]a_4 = a^2_3[/tex]

[tex]a_4 = 16^2[/tex]

[tex]a_4 = 256[/tex]

n =5

[tex]a_5 = a^2_{5-1[/tex]

[tex]a_5 = a^2_4[/tex]

[tex]a_5 = 256^2[/tex]

[tex]a_5 = 65536[/tex]

n = 6

[tex]a_6 = a^2_{6-1[/tex]

[tex]a_6 = a^2_{5[/tex]

[tex]a_6 = 65536^2[/tex]

[tex]a_6 = 4294967296[/tex]

[tex]a_1 = 2[/tex]   [tex]a_2 = 4[/tex]   [tex]a_3 = 16[/tex]   [tex]a_4 = 256[/tex]   [tex]a_5 = 65536[/tex]   [tex]a_6 = 4294967296[/tex]

Solving (c):

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] [tex]a_0=1[/tex]  ; ;  [tex]a_1=2[/tex]

[tex]a_1=2[/tex] ---- First term

n = 2

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_2=a_{2-1}+3a_{2-2}[/tex]

[tex]a_2=a_1+3a_0[/tex]

Substitute values for a1 and a0

[tex]a_2=2+3 * 1[/tex]

[tex]a_2=2+3[/tex]

[tex]a_2=5[/tex]

n = 3

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_3=a_{3-1}+3a_{3-2}[/tex]

[tex]a_3=a_{2}+3a_{1}[/tex]

[tex]a_3=5+3 * 2[/tex]

[tex]a_3=5+6[/tex]

[tex]a_3=11[/tex]

n = 4

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_4=a_{4-1}+3a_{4-2}[/tex]

[tex]a_4=a_{3}+3a_{2}[/tex]

[tex]a_4=11+3 * 5[/tex]

[tex]a_4=11+15[/tex]

[tex]a_4=26[/tex]

n = 5

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_5=a_{5-1}+3a_{5-2};[/tex]

[tex]a_5=a_{4}+3a_3[/tex]

[tex]a_5=26+3 * 11[/tex]

[tex]a_5=26+33[/tex]

[tex]a_5=59[/tex]

n = 6

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_6=a_{6-1}+3a_{6-2}[/tex]

[tex]a_6=a_{5}+3a_4[/tex]

[tex]a_6=59+3*26[/tex]

[tex]a_6=59+78[/tex]

[tex]a_6=137[/tex]

[tex]a_1=2[/tex]   [tex]a_2=5[/tex]   [tex]a_3=11[/tex]   [tex]a_4=26[/tex]   [tex]a_5=59[/tex]   [tex]a_6=137[/tex]

Solving (d):

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex];     [tex]a_0=1[/tex]; [tex]a_1=1[/tex]

[tex]a_1=1[/tex] --- First term

n = 2

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_2=2 * a_{2-1}+2^2a_{2-2}[/tex]

[tex]a_2=2 * a_1+4*a_0[/tex]

[tex]a_2=2 * 1+4*1[/tex]

[tex]a_2=2 +4[/tex]

[tex]a_2=6[/tex]

n = 3

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_3=3 * a_{3-1}+3^2 * a_{3-2}[/tex]

[tex]a_3=3 * a_{2}+9 * a_{1}[/tex]

[tex]a_3=3 * 6+9 * 1[/tex]

[tex]a_3=18+9[/tex]

[tex]a_3=27[/tex]

n = 4

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_4=4*a_{4-1}+4^2*a_{4-2}[/tex]

[tex]a_4=4*a_{3}+16*a_{2}[/tex]

[tex]a_4=4*27+16*6[/tex]

[tex]a_4=204[/tex]

n = 5

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_5=5 * a_{5-1}+5^2 * a_{5-2}[/tex]

[tex]a_5=5 * a_{4}+25 * a_{3}[/tex]

[tex]a_5=5 * 204+25 *27[/tex]

[tex]a_5=1695[/tex]

n = 6

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_6=6 * a_{6-1}+6^2*a_{6-2}[/tex]

[tex]a_6=6 * a_{5}+36*a_{4}[/tex]

[tex]a_6=6 * 1695+36*204[/tex]

[tex]a_6=17514[/tex]

[tex]a_1=1[/tex]   [tex]a_2=6[/tex]   [tex]a_3=27[/tex]   [tex]a_4=204[/tex]   [tex]a_5=1695[/tex]   [tex]a_6=17514[/tex]

Solving (e):

[tex]a_n= a_{n-1}+a_{n-3};\ a_0=1; a_1=2; a_2=0[/tex]

First term: [tex]a_1=2[/tex]

Second Term: [tex]a_2=0[/tex]

n = 3

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_3= a_{3-1}+a_{3-3}[/tex]

[tex]a_3= a_{2}+a_0[/tex]

[tex]a_3= 0+1[/tex]

[tex]a_3= 1[/tex]

n = 4

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_4= a_{4-1}+a_{4-3}[/tex]

[tex]a_4= a_{3}+a_{1}[/tex]

[tex]a_4= 1+2[/tex]

[tex]a_4=3[/tex]

n = 5

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_5= a_{5-1}+a_{5-3}[/tex]

[tex]a_5= a_{4}+a_{2}[/tex]

[tex]a_5= 3 + 0[/tex]

[tex]a_5= 3[/tex]

n = 6

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_6= a_{6-1}+a_{6-3}[/tex]

[tex]a_6= a_{5}+a_{3}[/tex]

[tex]a_6= 3 + 1[/tex]

[tex]a_6= 4[/tex]

[tex]a_1=2[/tex]   [tex]a_2=0[/tex]   [tex]a_3= 1[/tex]   [tex]a_4=3[/tex]   [tex]a_5= 3[/tex]   [tex]a_6= 4[/tex]