the sum of the squares of two consecutive negative even integers is 52. Find the integers. show all work!!!

1 answer:

Alex3 years ago

3 0

The two numbers are -4 and -6.

- two negatives make a positive. in this case -4×-4 equals 16 and -6×-6 equals 36. these are the squares

- 16+36 equals 52

please vote my answer branliest. Thanks!

You might be interested in

URGENTE, PERGUNTAS DE MARCAR:

Norma-Jean [14]

Answer:

2a: (c)

5o: (1, 3) and (1,1)

3a: (b)

1a: (d)

4o: (b)

Step-by-step explanation:

2a: the equation of a circle circumference needs to be transformable to the form $(x-x_C)^2 + (y - y_C)^2 = r^2$ where $C(x_C; y_C)$ is the center and r is the radius. (a) and (d) can’t be it because they contain non-zero factors on xy. (b) isn’t an equation.

5o: just put the given (x, y) into the equations and see if it holds. (2, 3) isn’t on the circumference of (1) because $2^2+3^2-2\cdot 2 -4\cdot 3 + 4 = 1 \neq 0$ , (3, 1) isn’t on it either because $3^2 + 1^2 - 2\cdot 3 - 4\cdot 1 + 4 = 4 \neq 0$ .

3a: calculate the value of the left-hand side term of the equation using (x, y) from the given point M. That’s the difference of square distance to the center to the square radius $r^2$ . Thus it’s 0 if the point is on the circumference, negative if inside and positive if outside. You get $3^2 + 4^2 - 6\cdot 3 -2\cdot 4 + 8 = 7$ , positive, so it’s outside the circle.

1a: see definition from 2a. Here, $r^2 = 9$ .

4o: insert the y from the straight line equation (r) (which can be equivalently transformed to $y = x+1$ ) into the circumference equation. If it yields no solution, that’s outside, it there’s exactly one solution, that’s a tangent and if there are two solutions, it’s a secant. $x^2+(x+1)^2+2x - 4(x+1) = x^2 + x^2 + 2x + 1 + 2x - 4x + 4 = 2x^2 + 5 = 0 \iff x^2 = 2.5 \iff x \in \{-\sqrt{2.5}, \sqrt{2.5}\}$ There are two solutions, so it’s a secant.