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gladu [14]
4 years ago
10

What are two integers whose sum is -2 and product is -80?

Mathematics
1 answer:
astra-53 [7]4 years ago
4 0

Answer:

We can write:

x + y = -2

xy = -80

We can rewrite the first equation as x = -y - 2 and then plug that into the second equation to get (-y-2) * y = -80 → -y² - 2y = -80 → y² + 2y - 80 = 0 → (y - 8)(y + 10) = 0 → y = 8, -10. Substituting these values into the first equation we get x = -10, 8 so the answer is (x₁, y₁) = (-10, 8) or (x₂, y₂) = (8, -10).

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ABCD is a rhombus. = 5.5 cm and = 4 cm. What is the area of the rhombus?
aleksandr82 [10.1K]
Assuming that the two given values refer to the two diagonals of the rhombus, its area can be calculated using the formula: A = 0.5*p*q, where, p and q are the two diagonals. Substituting the values to the formula, we get the area of rhombus ABCD to be equal to 11 cm^2. 
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The annual Gross Domestic Product (GDP) of a country is the value of all of the goods and services produced in the country durin
Vladimir [108]
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3 years ago
The sum of two numbers is 31 and their difference is 7
Diano4ka-milaya [45]

Answer:

24 and 7

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If you take 31-7=24, so 24+7=31 and 31-24=7

7 0
3 years ago
The part of the sphere x2 + y2 + z2 = 16 that lies above the cone z = x2 + y2 . (Enter your answer as a comma-separated list of
statuscvo [17]

Answer:

(x,y,z)=(ucos(v), usin(v), \sqrt{16-u^{2})

where 0\leq u\leq 2\sqrt{2} and 0\leq v\leq 2\pi

Step-by-step explanation:

Equation of a cone is z=\sqrt{x^{2} +y^{2}}

Equation of a paraboloid is z=x^{2} +y^{2}

I have parametrised cone here. Please note that equation for cone in the question, is actually a paraboloid.

Imagine a sphere of radius 4, centered at origin and intersecting a cone also centered at origin and height along positive z-axis, given by the equations

x^{2} +y^{2} +z^{2} = 16\\z=\sqrt{x^{2} +y^{2}}

where z\geq x^{2} +y^{2}

Solving for these two equations, and substituting for z in the equation of sphere, we get a circle of radius 2\sqrt{2} units.

The equation of intersecting circle is:

x^{2} +y^{2}=8

Now, according to question, parametrizing this region of circle using parameters u and v.

Consider cylindrical co-ordinates: (r,θ,z)

In cylindrical co-ordinates

(x, y, z)= (r cos(θ),  r sin(θ), z)

x^{2} +y^{2}= r^{2}

Eliminating z, and changing (r, θ)=(u,v)

For cone: x=ucos(v)

y= usin(v)

z=\sqrt{16-u^{2}

or (x,y,z)=(ucos(v), usin(v), \sqrt{16-u^{2})

where 0\leq u\leq 2\sqrt{2} and 0\leq v\leq 2\pi

6 0
3 years ago
2 2/3 +2 2/3 + 2 2/3 +2 2/3+5
Goryan [66]

Answer:

15 2/3

Step-by-step explanation:

5\frac{2}{3}  + 2 \frac{2}{3}  = 8

8 + 2 \frac{2}{3}  + 5

10 \frac{2}{3}  + 5 = 15 \frac{2}{3}

15 \frac{2}{3}

7 0
3 years ago
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