Answer:
dimo ba kayang answeran
Step-by-step explanation:
kaya muyan
Answer:
55, 125 and 125
Step-by-step explanation:
In this question, we are asked to find the other 3 angles in an intersection.
Please check attachment for what the diagram of the intersection might look like.
The intersection consists simply of 4 angles meeting at a point with the 2 angles on each side vertically opposite to the angles on the other side.
since one of the angles is 55, and we are having a straight line, the other angle would be 180-55 = 125.[ sum of angles on a straight line is 180]
These two angles are vertically opposite, this means that the other three angles are 55,125 and 125.
Answer:
The one on the top left
Step-by-step explanation:
What makes a function, a function is that the x coordinate (x,y), cannot be repeated in the tables. The y coordinate can though.
This is not a polynomial equation unless one of those is squared. As it stands x=-.833. If you can tell me which is squared I can help solve the polynomial.
Ok, that is usually notated as x^3 to be clear. I'll solve it now.
x^3-13x-12=0
Then use factor theorum to solve x^3-13x-12/x+1 =0
So you get one solution of x+1=0
x=-1
Then you have x^2-x-12 now you complete the square.
Take half of the x-term coefficient and square it. Add this value to both sides. In this example we have:
The x-term coefficient = −1
The half of the x-term coefficient = −1/2
After squaring we have (−1/2)2=1/4
When we add 1/4 to both sides we have:
x2−x+1/4=12+1/4
STEP 3: Simplify right side
x2−x+1/4=49/4
STEP 4: Write the perfect square on the left.
<span>(x−1/2)2=<span>49/4
</span></span>
STEP 5: Take the square root of both sides.
x−1/2=±√49/4
STEP 6: Solve for x.
<span>x=1/2±</span>√49/4
that is,
<span>x1=−3</span>
<span>x2=4</span>
<span>and the one from before </span>
<span>x=-1</span>
Answer:
The answer would be C
Step-by-step explanation:
You were going in the right direction but the problem was that you needed to subtract and not add. The formula should be y-y1 / x-x1 iirc. Hope this helped.
