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2 years ago

Find the 6th term of the arithmetic sequence -4, 0, 4, 8,

Mathematics

2 answers:

bezimeni [28]2 years ago

4 0

Answer:

Step-by-step explanation:

-4,0,4,8,12,16

Nastasia [14]2 years ago

3 0

A. because the arithmetic sequence increases by 4 and if u keep adding 4 till the 6th term, u will get 16.

4n-8

N being the Nth term

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Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch

enot [183]

Answer:

$y=\frac{-7t^2+22t-7}{7t-22}$

Step-by-step explanation:

We are given that

Initial value problem

$y'=(t+y)^2-1$ , y(3)=4

Substitute the value $z=t+y$

When t=3 and y=4 then

z=3+4=7

$y'=z^2-1$

Differentiate z w.r.t t

Then, we get

$\frac{dz}{dt}=1+y'$

$z'=1+z^2-1=z^2$

$z^{-2}dz=dt$

Integrate on both sides

$-\frac{1}{z}dz=t+C$

$z=-\frac{1}{t+C}$

Substitute t=3 and z=7

Then, we get

$7=-\frac{1}{3+C}$

$21+7C=-1$

$7C=-1-21=-22$

$C=-\frac{22}{7}$

Substitute the value of C then we get

$z=-\frac{1}{t-\frac{22}{7}}$

$z=\frac{-7}{7t-22}$

$y=z-t$

$y=\frac{-7}{7t-22}-t$

$y=\frac{-7-7t^2+22t}{7t-22}$

$y=\frac{-7t^2+22t-7}{7t-22}$

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2 years ago

One brand of cookies is sold in three different box sizes: small, medium, and large. The small box has half the volume of the me

lozanna [386]

Answer:

Step-by-step explanation:

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1 year ago

Ready Practice: GCF and LCM - Quiz - Level F Caylan is making baked goods for a charity bake sale. He places a tray of scones, a

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Which rational number is one tick mark to the right of Negative 6 Over 4 on the number line below?

murzikaleks [220]

Answer:

The number line is missing, but as we are know that the number marked in the number line is -6/4, i will guess that the ticks are spearated by fourts (the distance between each tick is 1/4).

Now, for the number at the right of -6/4, we should add the distance for one tick, this means that the number at the right is:

-6/4 + 1/4 = -5/4.

Now i will give some other examples:

Now, if the distance between ticks is 2/4, then the number at the right will be:

-6/4 + 2/4 = -4/4 = -1

Now, if the distance between ticks is 3/4, the the number at the right will be:

-6/4 + 3/4 = -3/4.

4 0

2 years ago

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = $\{ (a,b) \in R^2 \}$

d) R1ºR4 = $\{ (a,b) \in R^2 \}$

e) R1ºR5 = R1

f) R1ºR6 = $\{ (a,b) \in R^2 \}$

g) R2ºR3 = $\{ (a,b) \in R^2 \}$

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that ac. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

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2 years ago