How would I at least start out those two problems?
It looks to me like you already did. Now just follow through with it.
If the length of each side of the frame's inner dimension (i.e. the glass) is x, and the length of each side of the frame's outer dimension is x^2 - 3, then the area of the frame is:
A = (outer dimension)^2 - (inner dimension)^2
27 = (x^2 - 3)^2 - (x)^2
27 = x^4 - 6x^2 + 9 - x^2
27 = x^4 - 7x^2 + 9 <----- This is where you got to already.
0 = x^4 - 7x^2 - 18
0 = (x^2 - 9)(x^2 + 2)
Therefore, either x^2 = 9 or x^2 = -2... and you know it can't be the latter.
Thus, x^2 = 9 and therefore x = 3. The glass is 3" on each side.
<span>Each (outer) side of the frame is x^2 - 3 = 9 - 3 = 6" </span>
Answer:
it is 5.8
Step-by-step explanation:
Answer to your Question:
- The Minimum Value will be 0
and
- The Maximum Value will be 30
Other Answers to This Quiz Include:
"What will be the y-intercept in Hayato's Graph"
Answer: 30
"How will he label the first four tick marks on the X-Axis?"
Answer: 0.375, 0.75, 1.125, 1.5
"Which Graph is the first cycle of Hayato's Graph?"
Answer: Picture Below.
"At Which Time could the clown have dropped the bowling Pins?"
Answer: 3.75 s
This is the graph for y=-2(times)5^x
Hope this helps!
Answer:
Step-by-step explanation:
The drawing of the garage is a smaller representation of the actual garage. A garage is drawn using a scale of 2 2/3 inch = 6feet. Let us first convert 2 2/3 inch to improper fraction. It becomes 8/3 inch.
Let us determine the unit measurement.
If 6feet = 8/3 inch,
1 foot will be 8/3 ÷ 6 = 8/3×1/6
= 8/18 inches. Therefore
if the actual height is 68 ft , then, the height of the garage on the drawing will be
68 × 8/18 = 272/9 = 30 2/9 inches
