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BARSIC [14]
3 years ago
6

Write an equation in slope intercept form for the line that satisfies each set of conditions. Has y intercept -2 , parallel to t

he line that y=2x+4
Mathematics
1 answer:
Gwar [14]3 years ago
3 0
Y = 2x - 2

is the parallel line has a slip of 2 then your line has a slip of 2.
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The straight line y = mx +b passes through point C(−2, 3) and intersects the X-axis at point A, and the Y-axis at point B. Find
masya89 [10]
(-2,3) (2,1) -2-1=-3 3-2=1 so that means mx+-3
7 0
3 years ago
-3Ixl + 2x - 1, if x = -5<br><br> I don't get these types of problems :/
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5 0
2 years ago
Mathematicians 4: Who is Who?
larisa86 [58]

Answer:

a. x = -9 or x = -2

b. -5(x - 4)

c. x = -3 or x = 5

d. x = ±7

Step-by-step explanation:

a. First person;

y = x² + 11x + 18

y = x² + 9x + 2x + 18

y = x(x + 9) + 2(x + 9)

y = (x + 9)(x + 2)

y = x = -9 or x = -2

b. Second person;

y = -5x + 20

The common factor is 5.

y = -5(x - 4)

c. Third person;

y = x² - 2x - 15

y = x² - 5x + 3x - 15

y = x(x - 5) + 3(x - 5)

y = (x + 3)(x - 5)

y = x = -3 or x = 5

d. Fourth person;

y = x² - 49

Applying the difference of squares formula;

(a² - b²) = (a - b)(a + b)

y = x² - 49 = x² - 7² = (x - 7)(x + 7)

y = (x - 7)(x + 7)

y = x = ±7

6 0
2 years ago
What are the solutions to the nonlinear equations below y=4x x^2 + y^2=17
wolverine [178]

Answer:

x = 1 and x = -1

Step-by-step explanation:

Next time, please separate the equations with a comma or semicolon.  Thanks.

We can substitute 4x for y in the quadratic equation x^2 + y^2=17:

x² + (4x)²=17.

Then x² + 16x² = 17, or

17x² = 17, or x² = 1.  There are two solutions:  x = 1 and x = -1.

8 0
3 years ago
????⃗ (x,y)=(3x−4y)????⃗ +2x????⃗ F→(x,y)=(3x−4y)i→+2xj→ and ????C is the counter-clockwise oriented sector of a circle centered
Karolina [17]

By Green's theorem, the integral of \vec F along C is

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(2x)}{\partial x}-\frac{\partial(3x-4y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=6\iint_D\mathrm dx\,\mathrm dy

which is 6 times the area of D, the region with C as its boundary.

We can compute the integral by converting to polar coordinates, or simply recalling the formula for a circular sector from geometry: Given a sector with central angle \theta and radius r, the area A of the sector is proportional to the circle's overall area according to

\dfrac A{\frac\pi3\,\rm rad}=\dfrac{16\pi}{2\pi\,\rm rad}\implies A=\dfrac{8\pi}3

so that the value of the integral is

\dfrac{6\times8\pi}3=\boxed{16\pi}

6 0
3 years ago
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