The solution of the system of equation is the intersection point of the two quadratic equations, so we need to equate both equations, that is,
So, by moving the term -3x^3+20 to the left hand side, we have
Then, in order to solve this equation, we can apply the quadratic formula
In our case, a=5, b=-3 and c=-30. So we get
which gives
By substituting these points into one of the functions, we have
and
Then, by rounding these numbers to the nearest tenth, we have the following points:
Therefore, the answer is the last option
The area of the two triangular bases is
(total base area) = 2×(1/2)bh
= bh
= (10 m)(6 m) = 10·6 m² = 60 m²
The area of the three rectangles making up the remaining faces of the prism is
(total lateral area) = length×base + length×height + length×hypotenuse
= length×(base + height + hypotenuse)
= (16 m)(10 m + 6 m + 11.6 m)
= (16 m)(27.6 m) = 16·27.6 m² = 441.6 m²
Then the surface area of the prism is the sum of these
surface area = total base area + total lateral area
= 60 m² + 441.6² = 501.6 m²
When rounded to the nearest whole number, this is
A) 502 m²
Answer:
1. x=4+2y/3
2. x=3/4+y/2
Step-by-step explanation:
hope this helped :))
Answer:
idk!!!!! :c
Step-by-step explanation:
SORRY
One possible answer is 8H + 4L = 512
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Explanation:
- H = number of high resolution photos
- L = number of low resolution photos
H and L are placeholders for positive whole numbers, or 0 could replace either variable.
1 high resolution photo takes up 8 mb of space, so H of them take up 8H megabytes of space.
1 low resolution photo takes up 4 mb of space, so L of them take up 4L megabytes.
Overall, the two types of photos take up 8H+4L megabytes, which is equal to 512 since that's the max capacity. That leads to the equation 8H+4L = 512
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Extra info:
There are other ways to express the equation 8H+4L = 512. We could divide each part by 4 to get 2H+L = 128, but I think this equation loses its descriptive quality in a way. We no longer can see that each high resolution photo takes up 8 megabytes (instead we might mistakenly think only 2 megabytes are used per high resolution photo). A similar mistake may happen with the low resolution photos also.
The equation 2H+L = 128 can be rearranged into L = -2H+128 when solving for L.
