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3 years ago

Which equation represents the line that passes through the points (-3, 7) and (9,-1)? oy--3x+5 o y=-x-7 o y=zx-7 oy - 12 2x+5

Mathematics

1 answer:

nekit [7.7K]3 years ago

4 0

Answer:

use the slope equation (y2-y1) / (x2 -x1). plug in values and solve

(-1 -7) / (9 -(-3)) (subtracting a negative is the same as adding a positive)

-8 / 12 (simplify)

m = -2/3

then, plug in one of the points and the slope into the slope-point equation.

(y - y1) = m (x - x1) (the point-slope form)

y - 7 = -2/3( x - (-3)) (plug in the slope and point 1 values, then solve)

y - 7 = -2/3x - 2 (add 7 to both sides)

y = 2/3x +5 (the answer)

Step-by-step explanation:

I hope this helps :))

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PLEASE HELP WILL GIVE BRAINIEST

Usimov [2.4K]

Answer:

1. B

2. A $(y+1)=\frac{8}{9} (x+4)\\y+1=\frac{8}{9} x+\frac{8}{9} *4\\y+1=\frac{8}{9} x+\frac{32}{9}\\9*(y+1=\frac{8}{9} x+\frac{32}{9})\\9y+9=8x+32\\-8x+9y=32-9\\-8x+9y=23$

Step-by-step explanation:

The point slope form of a line is $(y-y_1)=m(x-x_1)$ where $x_1=-4\\y_1=-1$ . We write

$(y--1)=m(x--4)\\(y+1)=m(x+4)$

To find m, count the slope from each marked point on the graph. Notice one is 1/2 so we will count by halves. The slope is 8/9

This means B is the point slope form. To convert to the standard form, it must be written as $ax^2+by^2=c$ . We convert using inverse operations.

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3 years ago

Find the real solution<br><br> x^4-x^3+2x^2-4x-8=0

Law Incorporation [45]

Answer:

There's 2 solution, x= -1 and x= 2

Step-by-step explanation:

3 0

3 years ago

Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,

rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1$ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0, ⇒ dx = 0

y = 9 t, ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$\oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx$

= 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy$

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

$= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$ is conservative.

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3 years ago

What number is the divisor in the division problem below? 36 = 4= 9

Pani-rosa [81]

Umm isn’t it 36 I don’t know

7 0

3 years ago

Determine whether the sequence is arithmetic, geometric, both, or neither. 1, 4, 9, 16, 25, . . .

klemol [59]

Answer:

neither

Step-by-step explanation:

First differences are 3, 5, 7, 9, and the differences of these (2nd differences) are constant at 2. The degree of the polynomial function describing the sequence is equal to the number of the differences that are constant. Here, that is 2nd differences, so the sequences is described by a 2nd-degree (quadratic) polynomial.

It is not linear (arithmetic) or exponential (first differences have a common ratio).

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3 years ago

Which equation represents the line that passes through the points (-3, 7) and (9,-1)? oy--3x+5 o y=-x-7 o y=zx-7 oy - 12 2x+5​

Which equation represents the line that passes through the points (-3, 7) and (9,-1)? oy--3x+5 o y=-x-7 o y=zx-7 oy - 12 2x+5