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2 years ago

a chalk line is placed around the perimeter of the football field what is the length of this line in feet

Mathematics

1 answer:

masya89 [10]2 years ago

3 0

50 so thatbis the snswer

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PlZZZ HELP Which of the following is equal to

stealth61 [152]

Answer:

1st option

Step-by-step explanation:

4 $\frac{1}{3}$ ÷ 2 $\frac{3}{5}$ ( change mixed numbers into improper fractions )

= $\frac{13}{3}$ ÷ $\frac{13}{5}$

To perform the division

leave 1st fraction, change ÷ to × , turn 2nd fraction ' upside down'

= $\frac{13}{3}$ × $\frac{5}{13}$

7 0

1 year ago

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Evaluate the expression 6a - 2b when a = 5, b=7

melomori [17]

6a - 2b

= 6(5) - 2(7)

= 30 - 14

= 16 answer

6 0

2 years ago

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On one play, a football team gained 9 yards. On the next play, the football team received a penalty

mafiozo [28]

Answer:

If your wanting to find the difference in these numbers or trying to gind the sum of them... It would be -1

Step-by-step explanation:

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2 years ago

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A rectangular field with perimeter of 80m is to have an area of at least 380m² . Describe the possible lengths and of the field.

Kryger [21]

$let \: x \: be \: the \: length \\ let \: y \: be \: the \: width$

$perimeter = 2x + 2y \\ area = xy$

$2x + 2y = 80 \\ xy = 380 \\ \\ x + y = 40 \\ xy = 380 \\ \\ x = 40 - y \\ (40 - y)y = 380$

$40y - y {}^{2} = 380 \\ y {}^{2} - 40y + 380 = 0 \\ y {}^{2} - 40y + 400 = 20 \\ (y - 20) {}^{2} = 20 \\ y - 20 =± \sqrt{20} \\ y_{1}=20-√20≈15.52786\\ y_{2}=20+√20≈24.4721$

6 0

1 year ago

Value of the derivative of g(x)=8-10Cosx at 'x=0' is?

VLD [36.1K]

Answer:

g'(0) = 0

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Function Notation

Pre-Calculus

Unit Circle

Calculus

Derivatives
Derivative Notation
The derivative of a constant is equal to 0
Derivative Property: $\frac{d}{dx} [cf(x)] = c \cdot f'(x)$
Trig Derivative: $\frac{d}{dx} [cos(x)] = -sin(x)$

Step-by-step explanation:

Step 1: Define

g(x) = 8 - 10cos(x)

x = 0

Step 2: Differentiate

Differentiate [Trig]: g'(x) = 0 - 10[-sin(x)]
Simplify Derivative: g'(x) = 10sin(x)

Step 3: Evaluate

Substitute in x: g'(0) = 10sin(0)
Evaluate Trig: g'(0) = 10(0)
Multiply: g'(0) = 0

3 0

2 years ago