9514 1404 393
Answer:
- e ≈ 9.608
- D ≈ 46.5°
- F ≈ 38.5°
Step-by-step explanation:
The only given angle is between the two given sides, so the Law of Cosines must be used to find the side opposite the angle. It tells you ...
e² = d²+f² -2df·cos(E)
e² = 7² +6² -2·7·6·cos(95°) ≈ 92.321082
e ≈ 9.608
The remaining angles can be found from the law of sines.
sin(D)/d = sin(E)/e
D = arcsin(d/e·sin(E)) ≈ arcsin(7/9.608·sin(95°)) ≈ 46.5°
F = 180° -95° -46.5° = 38.5°
Answer:
Step-by-step explanation:
That 3 sitting outside there in that little "box" thing is a root/solution/zero of the polynomial. The numbers underneath the line are the coefficients of the depressed polynomial, which means that the polynomial is 1 degree lower than the degree with which we started. If we started with an x-squared, this degree is a single x, better known as linear (a line). Anyway, (x - 3) is a zero of the polynomial, which also means that it's a factor. So A applies. x = 3 is a root, so C applies. And F also because the depressed polynomial, the remainder, is 2x + 4.
To multiply a monomial by a polynomial, you distribute the monomial into each number of the polynomial so you would be using the distributive property.
Answer:
Hey No worries. Its very easy.
Step-by-step explanation:
Look,
Volume= 84 in.cube
Length=6 in.
Breadth=2 in.
therefore, Height= Volume/(Length*Breadth)
=84/(6*2)
=84/12
=7
Answer:
(1,10) That is the slope, if there is another line then the answer will be different!
Step-by-step explanation:
This is the slope-intercept formula, (y=mx+b)
Start by finding the y-intercept which is Positive 7 or plus 7 you should find it on the y-axis on the positive side.
Then you will need to identify the slope, the slope, in this case is 3, which needs to be in rise over run formula so 3 over 1.
To find a slope you must start at the y-intercept in your case is 7, then you must go up 3 and right 1.
Then identify the x and y axis.
If this is not the answer please provide this equation and another one to get them intersecting to get one point.
