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brilliants [131]
3 years ago
10

If y=2x-4 what is m and b using y=mx+b form?

Mathematics
2 answers:
stiks02 [169]3 years ago
6 0

Answer:

m➡ 2

b➡-4

Step-by-step explanation:

y=mx+b

m-slope

b-y-intercept

Scrat [10]3 years ago
3 0
The answer to that is X =2
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4, 3.7, 3.4, 3.1, ____, ____, ____ what are the next terms
Taya2010 [7]

Answer:

the next terms are 2.8 , 2.5 and 2.2

7 0
3 years ago
Show how the following sequences is an arithmetic or not <br>please help me ​
serious [3.7K]

Answer:

not arithmetic

Step-by-step explanation:

An arithmetic sequence has a common difference d between consecutive terms.

Given

T_{n} = 2n² - 1

Substitute in n = 1, 2, 3, 4 to generate the first 4 terms of the sequence

T_{1} = 2(1)² - 1 = 2 - 1 = 1

T_{2} = 2(2)² - 1 = 8 - 1 = 7

T_{3} = 2(3)² - 1 = 18 - 1 = 17

T_{4} = 2(4)² - 1 = 32 - 1 = 31

check the difference between consecutive terms

7 - 1 = 6

17 - 7 = 10

31 - 17 = 14

The differences are not common hence not an arithmetic sequence

5 0
4 years ago
Please answer this correctly I have to finish the sums by today
Naily [24]

Answer: i cant see it

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Which of the functions graphed below is continuous?
vaieri [72.5K]

Answer:

The answer is the second one

5 0
3 years ago
Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye
Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let \rho(x,y) be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

m=\int\limits \int\limits_D \rho(x,y) \, dA.

From the question, the given density function is \rho (x,y)=xye^{x+y}.

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx.

Since D is a rectangular region, we can apply Fubini's Theorem to get:

I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx.

Let the inner integral be: I_0=\int\limits^1_0xye^{x+y}dy, then

I=\int\limits^1_0(I_0)dx.

The inner integral is evaluated using integration by parts.

Let u=xy, the partial derivative of u wrt y is

\implies du=xdy

and

dv=\int\limits e^{x+y} dy, integrating wrt y, we obtain

v=\int\limits e^{x+y}

Recall the integration by parts formula:\int\limits udv=uv- \int\limits vdu

This implies that:

\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy

\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}

I_0=\int\limits^1_0 xye^{x+y}dy

We substitute the limits of integration and evaluate to get:

I_0=xe^x

This implies that:

I=\int\limits^1_0(xe^x)dx.

Or

I=\int\limits^1_0xe^xdx.

We again apply integration by parts formula to get:

\int\limits xe^xdx=e^x(x-1).

I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1).

I=\int\limits^1_0xe^xdx=0-1(0-1).

I=\int\limits^1_0xe^xdx=0-1(-1)=1.

No unit is given, therefore the mass of the lamina is 1.

3 0
3 years ago
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