Answer:
y = -x -2
Step-by-step explanation:
The graph has a slope of -1 and a y-intercept of -2.
The equation is therefore
y = -x -2
The square box is enough to fit the pizza with a diameter of 10 inches inside. Since the area of the square box is more than the area of the pizza, the pizza fits easily in the square box.
<h3>What is the area of the circle and the square?</h3>
The area of the circle is
Ac = πr² = πd²/4 sq. units
Where r is the radius and d is the diameter of the circle.
The area of the square is given by
As = s² sq. units
Where s is the length of the side of a square.
<h3>Calculation:</h3>
It is given that a pizza(in a circular shape) with a diameter d = 10 in is to be placed in a square box of the same length as the diameter of the pizza.
So,
The area of pizza is
Ap = Ac = πd²/4 sq. units
= π(10)²/4
= 25π
= 78.54 sq. in
Then, the area of the square box with the length same as the diameter of the pizza is,
As = d²
= 10²
= 100 sq. in
Since the area of the square is more than the area of the pizza (100 sq. inch > 78.54 sq. inch), the pizza easily fits into the square box.
Learn more about the area of a circle here:
brainly.com/question/15673093
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Answer: x axis is time in hours, y axis is water level in centimeters
I can't see the rest of the problem since you didnt show it
Step-by-step explanation:
Answer:
170
Step-by-step explanation:
Ann, Ben, and Cindy were eating strawberries.
The ratio of the numbers of berries they ate is 5:5:7.
If Cindy ate 30 strawberries less than Ann and Ben together,
find:
what is the total number of strawberries the three of them ate?
solution:
add the ratios 5 + 5 + 7 = 17
since Cindy ate 30, less than Ann and Ben together so, the equation is
7x = 5x + 5x - 30
7x - 10x = -30
x = 30/3
x = 10
ann 5 x 10 = 50
ben 5 x 10 = 50
cindy 7 x 10 = 70
total = 170
Answer:
Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
Step-by-step explanation:
Given the figure with dimensions. we have to find the area of given figure.
Area of figure=ar(1)+ar(2)+ar(3)
Area of region 1 = ar(ANGI)+ar(AIB)
![=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha](https://tex.z-dn.net/?f=%3DL%5Ctimes%20B%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height%5C%5C%5C%5C%3D%5B1500%5Ctimes%20%285000-2000-1500%29%5D%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%283000-1500%29%5Ctimes%20%285000-2000-1500%29%5C%5C%5C%5C%3D3375000m%5E2%3D337.5ha)
Area of region 2 = ar(DHBC)
Area of region 3 = ar(GFEH)
Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha
=987.5 ha
Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.
Let the fencing be done through x m downward from B which divides the two into equal area.
⇒ Area of upper part above fencing=Area of lower part below fencing
⇒
Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
