MATHEMATICS HIGH SCHOOL

The numbers on a football field indicate 10 yard increments. You walk around the perimeter of a football field between the pylons. You walk a distance of 306 and 2/3 yards. Find the area and perimeter of the indicated regions. Write your answers as mixed numbers, if necessary. One end zone

Answers

Answer 1

Answer:

Answer: 126.67 is the perimeter walked

Step-by-step explanation:

Given data:

Distance walked on the field = 306 2/3 yards.

Increments on the field = 10 yards.

Solution:

We know a pitch has 4 sides let make it A, B,C,D = 306.67yards.

each has an increment of 10

= 10 x 10 + 10 x 10 = 200

Therefore

= 306.67 yards – 200

= 106.67 yards.

For both end zones Perimeter

= 106.67 + 10 + 10

= 126.67 yards

Answer 2

Answer:

Final answer:

The area of the end zone is 6400 square yards and the perimeter is 326 and 2/3 yards.

Explanation:

To find the area and perimeter of the end zone of a football field, we need to know the dimensions of the end zone. If the end zone has a width of 53 and 1/3 yards, we multiply the width by the length of the field, which is 120 yards, to find the area. So the area of the end zone is (53 and 1/3) x 120 = 6400 square yards. The perimeter is found by adding the dimensions together, so the perimeter of the end zone is (2 x 53 and 1/3) + (2 x 120) = 326 and 2/3 yards.

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Using the graph of f(x) = log10x below, approximate the value of y in the equation 10y = 6

Answers

The given function is

f(x)= log 10 x

Taking base as 10

$y=\log_(10)10 +\log_(10)x$ →→Using the property of, log a b= log a + log b

→y = 1 + $\log_(10)x$ as, $\log_(10)10=1$

we have to approximate the value of y in the equation

→10 y = 6

→y= $(6)/(10)$

$(6)/(10)$ = 1 + $\log_(10)x$

$\log_(10)x$ = $(-4)/(10)$

x= $(10)^{(-4)/(10)}$ = $(10)^(-0.4)$ =0.398

Solving graphically,

we get , x= 0.398 for , y= $(6)/(10)$ =.6

f(x) = ㏒(10x)
f(x) = ㏒₁₀(10x)

10y = 6
10 10
y = 0.6

f(x) = ㏒₁₀(10x)
0.6 = ㏒₁₀(10x)
$10^(0.6) = 10x$
$10^{(3)/(5)} = 10x$
$\sqrt[5]{10^(3)} = 10x$
$\sqrt[5]{1000} = 10x$
$\frac{\sqrt[5]{1000}}{10} = x$
$(3.98)/(10) \approx x$
$0.398 \approx x$

79/10 as a mixed number

Answers

7 and 9/10

Working;
10 goes into 70 7 times. The remainder is 9.

What is the ratio 18:24 in its simplest form

Answers

Answer:

3:4

Step-by-step explanation:

18:24

$(18)/(24)$

$(18 / 6)/(24 / 6)$

$(3)/(4)$

3:4

What's the value of log2 (1/8) ?

Answers

$log_aa^p=p\n------\n\nlog_2 ( (1)/(8) ) =log_28^(-1)=log_2(2^3)^(-1)=log_22^(-3)=-3\n\nAns.\ -3$

The correct answer for the value $log_(2)(1)/(8)$ is equal to -3.

What is Logarithm?

In mathematics, Logarithms are defined a way of expressing exponents. A logarithm is defined as the power to which a number must be raised to get some other values.

The expression for logarithm of a number is written as ㏒ₓb = y.

Properties of Logarithm:

$log_(x) (x^n) = n$

The value of $log_(2)(1)/(8)$ can be calculated by recognizing that $(1)/(8)$ is equal to 2 raised to the power of -3.

$log_(2)(1)/(8)$ = $log_(2)(2^(-3))$

From the property of logarithm:

$log_(2)(2^(-3))$

= -3

The value of $log_(2)(1)/(8)$ is -3.

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Does the following system have a unique solution? Why?

3x+2y= 9
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Answers

Yes, the determinant of the coefficient matrix is 2.

A weight is attached to to a spring that is fixed to the floor. The equation h=7cos (pi/3 t) models the height, h, in centimeters after t seconds of the weight being stretched and released. Solve for the solution of t and find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round to the nearest hundreth

Answers

Answer:

$t=(3)/(\pi)\left(\cos ^(-1)(h)/(7)\right)$

Step-by-step explanation:

The given function is,

$h=7\cos \left((\pi)/(3)t\right)$

$\Rightarrow \cos \left((\pi)/(3)t\right)=(h)/(7)$

$\Rightarrow (\pi)/(3)t=\cos ^(-1)(h)/(7)$

$\Rightarrow t=(3)/(\pi)\left(\cos ^(-1)(h)/(7)\right)$

This is the solution of t.

Now we have to find t when h is 1 cm, 3 cm and 5 cm.

When h=1

$t=(3)/(\pi)\left(\cos ^(-1)(1)/(7)\right)=78.10\ s$

When h=3

$t=(3)/(\pi)\left(\cos ^(-1)(3)/(7)\right)=61.71\ s$

When h=5

$t=(3)/(\pi)\left(\cos ^(-1)(5)/(7)\right)=42.41\ s$

lets solve for t in the height equation:

h = 7 cos (pi/3 t)
cos (pi/3 t) = h/7
pi/3 t = cos^-1 (h/7)

t = (3/pi) cos^-1 (h/7)

then we substitute:

t = (3/pi) cos^-1 (1/7)
t = 78.1 s

t = (3/pi) cos^-1 (3/7)
t= 61.71

t = (3/pi) cos^-1 (5/7)
t = 42.41

those are the times in seconds respectively.

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