A circle is drawn within a square as shown.



What is the best approximation for the area of the shaded region?

Use 3.14 to approximate pi.


21.5 cm²

78.5 cm²

100 cm²

178.5 cm2

In this problem, you are asked to find the area of the trapezoid. The formula in finding the area of the trapezoid is:

A = [(a + b)/2] x h

Where a = base 1

             b = base 2

             h = height

Substituting the given measurements to the formula:

A = [(1.7 m + 6.7 m) / 2] x 5 m

A = (8.4 m / 2) x 5 m

A = 4.2 m x 5 m

A = 21 m^2

Therefore, the area of the trapezoid is 21 square meters.

In triangle ADC and BCD, AB = DC (opposite sides of a rectangle are congruent)

In triangle ADC and BCD, AB = DC (opposite sides of a rectangle are parallel)

In triangle ADC and BCD, AD = BC (opposite sides of a rectangle are congruent)

In triangle ADC and BCD, AD = BC (opposite sides of a rectangle are parallel)

are the statments

The speed of the debris is 79.4 ft/s

Since there is an initial velocity of 160 feet per second, height is 300feet

So, we applied the below formula

[tex]v_({final})^2 = v_({initial})^2 + 2ah\\\\ v_({fina})l^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft)\\\\ = 6310 ft^2/s^2\\\\ = (6310 ft^2/s^2)^{1/2}\\\\ = 79.4 ft/s.[/tex]

Therefore, The speed of the debris is 79.4 ft/s

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We want to answer different things about an exponential decay, the answers are:

So we know that the original price of the computer is $1250 and the value decays by 10% each year.

So in year 1, the new value of the computer will be:

V = $1250*(1 - 0.1) = $1250*0.9

On year 2, the new value will be:

V = ( $1250*0.9)*0.9 = $1250*(0.9)^2

And so on.

a) This sequence is a geometric sequence because each term is a constant times the previous term, where the constant is 0.9.

[tex]A_1 = 1250\\A_2 = 0.9*1250 = 1125\\A_3 = 0.9^2*1250 = 0.9*1125 = 1012.5 \\...[/tex]

b) The explicit formula for a geometric sequence is:

[tex]A_n = k^{n-1}*A_1[/tex]

where:

k is the constant, in this case, is 0.9

A1 is the first term of the sequence, in this case, is 1250

Then we have:

[tex]A_n = (0.9)^{n-1}*1250[/tex]

c) Here we just need to replace n by 6 in the above formula:

[tex]A_6 = (0.9)^5*1250 = 738.1125[/tex]

This means that the price of the computer in the 6th year is $738.10

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The factorised form of 6m + 18 is given by the equation: 6(m + 3).

Given :

Equation - 6m + 18

Solution :

Factorisation is the method of diminishing the bracket of a quadractic condition, rather than growing the bracket and changing over the condition to a item of variables which cannot be decreased assist.

To factorise the equation 6m + 18 we have to follow some steps:

In first step we have to write 18 = [tex]6\times 3[/tex] so the equation becomes:

[tex]\rm = 6m + (6\times3)[/tex]  ----- (1)

In second step we have to take 6 common from equation (1):

[tex]\rm=6(m+3)[/tex]

So the factorised form of 6m + 18 is given by the equation: 6(m + 3)

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Answer:

His rate of discount=10%

Step-by-step explanation:

Step 1: Get the discount amount

Discount amount=List price-sale price

where;

List price=$800

Sale price=$720

replacing;

Discount amount=(800-720)=80

Discount amount=$80

Step 2: Calculate discount rate

The discount rate can be calculated as;

discount rate=(discount/List price)×100

where;

discount=$80

List price=$800

replacing;

discount rate=(80/800)×100=10%

discount rate=10%

Answer:

d. $10,234.80

Step-by-step explanation:

we are given

The equipment will cost $26,000

so, Amount is $26000

A=26000

should be invested today at 12%

r=12%=0.12

It is compounded semiannually

so, n=2

t=8

now, we can use formula

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]

now, we can plug values

[tex]26000=P(1+\frac{0.12}{2})^{2\times 8}[/tex]

now, we can solve for P

[tex]P=10234.80338[/tex]

Answer : The equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

Explanation :

Half-life = 5.27 years

Formula used :

[tex]a=\frac{a_o}{2^n}[/tex]        ............(1)

where,

a = amount of reactant left after n-half lives

[tex]a_o[/tex] = Initial amount of the reactant

n = number of half lives

And as we know that,

[tex]n=\frac{t}{t_{1/2}}[/tex]        ..........(2)

where,

t = time

[tex]t_{1/2}[/tex] = half-life  = 5.27 years

Now equating the value of 'n' from (2) to (1), we get:

[tex]a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}[/tex]       ...........(3)

[tex]a=\frac{a_o}{2^{(\frac{t}{5.27})}}[/tex]

[tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

Therefore, the equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)^(t/h), where t is the number of years and h is the half-life of the isotope.

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)(t/h), where t is the number of years and h is the half-life of the isotope.

For example, for the radioisotope cobalt-60 with a half-life of 5.27 years, you can use the equation Percent remaining = (100) x (1/2)(t/5.27) to determine the percent remaining after t years.

Let's say you want to find the percent remaining after 10 years, you would substitute t with 10 in the equation, like this: Percent remaining = (100) x (1/2)(10/5.27). Simplify the equation to find the answer.

Final answer:

The series 5+25+125+625+3125+15625 can be rewritten in sigma notation as Σ 5*5^(n-1) from n=1 to 6, representing a geometric series with a common ratio of 5 and 6 terms.

Explanation:

To rewrite the series 5+25+125+625+3125+15625 using sigma notation, we first need to identify the pattern of the series. We notice that each term is 5 times the previous term, which is a characteristic of a geometric series. A geometric series can be expressed in sigma notation as Σ a*r^(n-1) from n=1 to N, where a is the first term, r is the common ratio between terms, and N is the number of terms.

In this series, a = 5, and r = 5. The series has 6 terms, so N = 6. Therefore, the series in sigma notation is Σ 5*5^(n-1) from n=1 to 6.

This detailed answer covers evaluating functions at specified values, identifying the y-intercept of a function's graph, determining asymptotes, identifying points of maximum growth rate in logistic functions, and intervals where the function's growth rate is decreasing, all critical concepts in understanding mathematical functions in high school.

The questions provided are focused on evaluating specific values of functions, understanding the y-intercept of a graph, determining asymptotes, identifying points of maximum growth rate, and assessing intervals of decreasing growth rate for various exponential and logistic functions. Each of these concepts plays a crucial role in understanding the behaviors and characteristics of different types of mathematical functions, which are fundamental in high school mathematics.