3) Nancy is 22 years older than her son Roger. In 8 years Nancy will be 3 times as old as Roger is in 2 years.

Find one counter example to show that the conjecture is false. angle 1 and angle 2 are supplementary, so one of the angles is a

Mrac [35]

Answer:

B. m ∠ 1 = 90° and m ∠ 2 = 90°

Step-by-step explanation:

For most situations, the conjecture would probably be true, but there is one exception that makes this statement false.

When two right angles are supplementary, none of them is acute.

For an angle to be acute it needs to be lesser than 90°, and for a pair of angles to be supplementary they should add up to exactly 180°.

With a pair of right angles (90° each), their sum adds up to 180° but neither of them are acute.

Therefore, the answer is B. m ∠ 1 = 90° and m ∠ 2 = 90°

4 0

3 years ago

Solve 3x-2y=15 for y?

Nikitich [7]

Answer:

y=3x-15

Step-by-step explanation:

3x-2y=15

+2y +2y

3x=15+2y

-15 -15

y=3x-15

7 0

3 years ago

Using a directrix of y = −2 and a focus of (1, 6), what quadratic function is created? Question 3 options:

deff fn [24]

If your directrix is a "y=" line, that means that the parabola opens either upwards or downwards (as opposed to the left or the right). Because it is in the character of a parabola to "hug" the focus, our parabola opens upwards. The vertex of a parabola sits exactly halfway between the directrix and the focus. Since our directrix is at y = -2 and the focus is at (1, 6) AND the parabola opens upward, the vertex is going to sit on the main transversal, which is also the "line" the focus sits on. The focus is on the line x = 1, so the vertex will also have that x coordinate. Halfway between the y points of the directrix and the focus, -2 and 6, respectively, is the y value of 2. So the vertex sits at (1, 2). The formula for this type of parabola is $(x-h)^2=4p(y-k)$ where h and k are the coordinates of the vertex and p is the DISTANCE that the focus is from the vertex. Our focus is 4 units from the vertex, so p = 4. Filling in our h, k, and p: $(x-1)^2=4(4)(y-2)$ . Simplifying a bit gives us $(x-1)^2=16(y-2)$ . We can begin to isolate the y by dividing both sides by 16 to get $\frac{1}{16}(x-1)^2=y-2$ . Then we can add 2 to both sides to get the final equation $f(x)=\frac{1}{16}(x-1)^2+2$ , choice 4 from above.