Cos A/1-tan A + sin A/1-cot A =sin A + cos A
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Related Questions
A surveyor measures the angle of elevation to the top of a building to be 70 degrees. The surveyor then walks 50 ft farther from the base of the tower and measures the angle of elevation to be 50 degrees. The surveyor's angle-measuring device 5.5 ft from the ground. How tall is the building, to the nearest foot?
0.09502 to 2 decimal places
Let f(x) = 3x2 – x 2 and g(x) = 5x2 – 1. Find f(g(x)). Show each step of your work.
In a sale game boys are reduced by two-fifths what is the sale price if the original price was 50
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The postal service delivered:
7.14 × 1010 pieces of mail in December
3.21 × 1010 pieces of mail in January
Total number of mail delivered is;
(7.14 + 3.21) × 1010
= 10.35 × 1010
= 1.035 × 10100 (in one decimal point form)
Answer:
The correct answer for this problem is
d) 1.035 x 10^11
Step-by-step explanation:
I know this because I just did the test
b) Determine x so that the volume of the box is at least 450 cubic inches.
c) Determine x so that the volume of the box is maximum.
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The volume of the box as a function of x V(x) = x ( 60 -2x )( 15-2x )
The volume of the box as a function of x inches 0.55 inches ≤ x ≤ 6.79
The volume of the box is maximum x ≥ 6.79 inches
Given ,
The box with no top that is to be made by removing squares of width x
The corners of a 15-in by 60-in piece of cardboard.
- Volume can be represented by this function below
V(x) = x ( 60 -2x )( 15-2x )
Where : x = height , ( 60 - 2x ) = length , ( 15 -2x ) = width
The volume of the box as a function of x is V(x) = x ( 60 -2x )( 15-2x )
- To determine x so that,
The volume of the box ≥ 450 inches
V(x) = x ( 60 -2x )( 15-2x )
The volume of the box is at least 450 cubic inches.0.55 inches ≤ x ≤ 6.79 inches
- The value of x for which volume of the box is maximum will be x ≥ 6.79 inches.
For more information about Volume of the square click the link given below
Answer:
a) V(x) = x ( 60 -2x )( 15-2x )
b) 0.55 inches ≤ x ≤ 6.79 inches
c) x ≥ 6.79 inches
Step-by-step explanation:
Given data:
No top, cardboard dimensions ; 15-in by 60-in
a) A function for the volume of the box as a function of x the Volume can be represented by this function below
= V(x) = x ( 60 -2x )( 15-2x )
where : x = height , ( 60 - 2x ) = length , ( 15 -2x ) = width
b) determine x so that the volume of the box ≥ 450 inches
450 = x( 60 - 2x ) ( 15 -2x ) ( solving the equation )
0.55 inches ≤ x ≤ 6.79 inches
c ) The value of x for which volume of the box is maximum
will be x ≥ 6.79 inches
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you have
To solve...you must complete the square...
use
x=1(+/-)
Thus, your answer.
(Or, x≈3.23606797, -1.236067967)
A.Which is the best comparison of the estimates and the actual product?
B.Both estimates will be less than the original product.
C. Both estimates will be greater than the original product.
Katie’s estimate will be less than the actual product, and Amaya’s estimate will be greater than the actual product.
D. Katie’s estimate will be greater than the actual product, and Amaya’s estimate will be less than the actual product.
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Answer:
Amaya's estimate is the closer one.
D. Katie’s estimate will be greater than the actual product, and Amaya’s estimate will be less than the actual product
Step-by-step explanation:
The original expression is 89.1 x 9.3. Katie changed the 89.1 to 99 (difference of 9.9) and the 9.3 to 10 (difference of 0.7). On the other hand, Amaya changed the 89.1 to 89 (0.1 difference) and the 9.3 to 9 (0.3 difference).
Therefore the best comparison would be Amaya's estimate since her numbers are closer to the original expression.
Katie's estimate will be greater than the actual product because she rounded up the numbers and Amaya's estimate will be less than the actual product because she rounded down the numbers.
Choice D), since Katie went way over the decimals, and Amaya went slightly under to the closest number
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69+21=90
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The perimeter of the garden is the sum of the two sides, and it is equal to 48 cm:
The area of the garden is the product between the two sides, and it is equal to
From (1), we find
And by substituting this into (2)
which gives two solutions:
So, the width of the garden is