This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get
i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:
where is a compact way to indicate both solutions and , while are the coefficients of the quadratic equation, i.e. we consider the polynomial .
So, in your case, we have
Plug those values into the formula to get
So, the two solutions are
Answer: The correct congruence statement is .
Explanation:
It is given that A triangle Q D J. The base D J is horizontal and side Q D is vertical. Another triangle M C W is made. The base C W and side M C are neither horizontal nor vertical. Triangle M C W is to the right of triangle Q D J.
The sides Q D and M C are labeled with a single tick mark. The sides D J and C W are labeled with a double tick mark. The sides Q J and M W are labeled with a triple tick mark.
Draw two triangles according to the given information.
From the figure it is noticed that
So by SSS rule of congruence we can say that .
It would be 100. Because let's say you had to round 25 to the nearest 10 it would be 30 because everything ending with 5 (example: 5 to nearest 10 would go to 10 65 to the nearest ten would go to 70) but anything lower then 5 goes to the lower number.