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2 years ago

A student says that the slope of the line for the equation y=20-15x is 20 and the y-intercept is 15. find and correct the error.

Mathematics

1 answer:

Sav [38]2 years ago

8 0

I dont really know but i think the intercept is 20

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What is the solution to the system of equations graphed below?

photoshop1234 [79]

A because that is where they cross

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3 years ago

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Because of variability in the manufacturing process, the actual yielding point of a sample of mild steel subjected to increasing

Alex73 [517]

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modify tge production

Step-by-step explanation:

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3 years ago

Marcella incarica Dora di aiutarla a suddividere 35 perle tra le sue tre figlie. Alla figlia maggiore vuole regalare la metà, al

Blizzard [7]

Answer:

La hija mayor recibirá 17.5 perlas, la hija del medio recibirá 11.66 perlas, y la hija menor recibirá 3.88 perlas.

Step-by-step explanation:

Dado que Marcella le pide a Dora que la ayude a dividir 35 perlas entre sus tres hijas, y quiere darle la mitad a la hija mayor, la tercera parte a la del medio y la novena a la menor y, además, Marcella quiere darle una perla a Dora como recompensa, para determinar cuántas perlas recibirá cada hija se debe realizar el siguiente cálculo:

35 / 2 = Mayor = 17.5 perlas

35 / 3 = Medio = 11.66 perlas

35 / 9 = Menor = 3.88 perlas

Así, la hija mayor recibirá 17.5 perlas, la hija del medio recibirá 11.66 perlas, y la hija menor recibirá 3.88 perlas.

5 0

2 years ago

Gradient from 2 - 1/2y = 0

kirill115 [55]

Given :

2 - 1/2y = 0

To find :-

The gradient form .

Solution :-

The gradient form/ slope intercept form

y = mx + c ,

On converting ,

2 - 1/2y = 0
1/2y = 2
y = 2*2
y = 4
y = 0x + 4

4 0

2 years ago

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago