Answer:
Step-by-step explanation:
4x+6<= -54
4x <= -60
x <= -15 [The point -15 is included, so it must be a solid dot]
----------------
5x + 7 > -18
5x > -25
x > -5 [The point -5 is NOT included, so it should be an empty dot.]
Solution C seems the best.
<span> b4 + b3 - 45b2 - 27b + 25
—————————————————————————
b2 </span>
Answer:
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola
y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?
Width =
Height =
Width =√10 and Height 
Step-by-step explanation:
Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)
are (h,k) and (-h,k).
Hence, the area of the rectangle will be (h + h) × k
Therefore, A = h²k ..... (2).
Now, from equation (1) we can write k = 5 - h² ....... (3)
So, from equation (2), we can write
![A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}](https://tex.z-dn.net/?f=A%20%3Dh%5E%7B2%7D%20%5B5-h%5E%7B2%7D%20%5D%3D5h%5E%7B2%7D%20-h%5E%7B4%7D)
For, A to be greatest ,

⇒ ![h[10-4h^{2} ]=0](https://tex.z-dn.net/?f=h%5B10-4h%5E%7B2%7D%20%5D%3D0)
⇒ 
⇒ 
Therefore, from equation (3), k = 5 - h²
⇒ 
Hence,
Width = 2h =√10 and
Height = 
Answer:
JM = 20
Step-by-step explanation:
Find the value of JM
Using similar triangles;
MJ/HN = GJ/HK
GJ+JM/HN = GJ/HK
15+JM/12+16 = 15/12
15+JM/28 = 15/12
Cross multiply
12(15+JM) = 15*28
180+12JM = 420
12JM = 420 - 180
12JM = 240
JM = 240/12
JM = 20
<em>Hence the length of JM is 20</em>