The cost of each bag of soil is $ 16.25 and the cost of each bag of fertilizer is $ 12.99.
<h3>What is the linear system?</h3>
A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.
Mr. Ellis has started a vegetable garden.
He bought 15 bags of soil and 3 bags of fertilizer for 282.72.
He realized he did not have enough supplies so he bought another 5 bags of soil and 2 bags of fertilizer for 107.23.
Let the cost of each bag of soil be x and the cost of a bag of fertilizer be y.
Then we have the linear equations,
15x + 3y = 282.72
5x + 2y = 107.23
On solving the equations 1 and 2, we have
x = 16.25 and y = 12.99
The cost of each bag of soil is $ 16.25 and the cost of each bag of fertilizer is $ 12.99.
More about the linear system link is given below.
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The answer is 7/5 th. You can add fractions without first having a common denominator so you would change the 8's by multiplying it by 5 to get 40 and you would multiply the 5 by 8 to get 40 as well. And the rule whatever you do to the bottom you have to do to the top so whatever you multiplied 8 by do the same for the numerator and whatever you multiplied 5 by you do the same for the numerator. So your new equation would be (75/40+16/40)+(-35/40). you would then add the numerator's across to get 56 and the denominator would stay the same so the answer is 56/40. But you have to now simplify 56/40 down to 14/10 and then once more to 7/5 which is the final answer.
Given :
C, D, and E are col-linear, CE = 15.8 centimetres, and DE= 3.5 centimetres.
To Find :
Two possible lengths for CD.
Solution :
Their are two cases :
1)
When D is in between C and E .
. . .
C D E
Here, CD = CE - DE
CD = 15.8 - 3.5 cm
CD = 12.3 cm
2)
When E is in between D and C.
. . .
D E C
Here, CD = CE + DE
CD = 15.8 + 3.5 cm
CD = 19.3 cm
Hence, this is the required solution.
Answer:
400 units³
Step-by-step explanation:
The volume (V) of the square pyramid is
V = 
area of base × height (h)
where h is the perpendicular height.
Consider the right triangle formed by a segment from the vertex to the midpoint of the base and the slant height ( the hypotenuse )
Using Pythagoras' identity on the right triangle
h² + 5² = 13²
h² + 25 = 169 ( subtract 25 from both sides )
h² = 144 ( take the square root of both sides )
h = 
= 12
Area of square base = 10² = 100, hence
V = × 100 × 12 = 4 × 100 = 400
