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Bess [88]
2 years ago
13

Quadrilateral $abcd$ is a trapezoid with $ab$ parallel to $cd$. we know $ab = 20$ and $cd = 12$. what is the ratio of the area o

f triangle $acb$ to the area of the trapezoid $abcd$? express your answer as a common fraction.

Mathematics
1 answer:
alekssr [168]2 years ago
4 0
Draw a diagram to illustrate the problem as shown below.

The area of triangle acb is
A₁ = (1/2)*20*h = 10h

The area of trapezoid abcd is
A₂ = (1/2)*(20+12)*h = 16h

The ratio A₁/A₂ is
A₁/A₂ = (10h)/(16) = 5/8

Answer:
The ratio of triangle acb to the area of trapezoid abcd is 5/8.

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GrogVix [38]

The rate to convert yards to meters is

1 \text{ yard}=0.9144\text{ meters}

So, if you multiply both sides by 8, you get

8 \text{ yards} = 8\times 0.9144 = 7.3152 \text{ meters}

To round the answer to the nearest tenth, we need to choose between 7.3 and 7.4. Since the hundreths digit is 1, which is less than 5, 7.3 is nearest to 7.3152 than 7.4, so the correct answer is 7.3

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2 years ago
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Jason collects baseball cards. He bought one card for $25. Its current value is represented by the expression 25(1.02)t, where t
ki77a [65]
Is Jason’s baseball card going up by 2% every year for example

Jason’s baseball card after 1 year is
25×(1.02)^(1)=25.5
It increased by 0.5 in amount after one year
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2 years ago
T.J. invests $8,100 in an account
Illusion [34]

Answer:

$13,177.97

Step-by-step explanation:

Compound Interest Formula

A=P(1+\frac{r}{n} )^{nt}

A = final amount

P = initial investment

r = interest rate

n = number of times the rate is applied during a time period

t = number of times the time period has elapsed

A=8100(1+\frac{.072}{1} )^{(1)(7)} \\A=8100(1.072)^{7}\\ A=8100*(1.626909883)\\A=13177.97006

6 0
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True or false? If LM=MP, then M must be the midpoint of LP
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2 years ago
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Conservation of Species A certain species of turtle faces extinction because dealers collect truckloads of turtle eggs to be sol
JulijaS [17]

Answer:

The turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

Ten years after the conservation measures are implemented the population will be 3260 turtles.

Step-by-step explanation:

To find the rate of growth of the turtle population at any time <em>t</em> you need to find N'(t)

\frac{d}{dt}N(t)=\frac{d}{dt}(2t^3+3t^2-4t+1000)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dt}(2t^3)+\frac{d}{dt}(3t^2)-\frac{d}{dt}(4t)+\frac{d}{dt}(1000)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\N'(t)=6t^2+6t-4

In particular, when t = 2 and t = 6, we have

N'(2)=6(2)^2+6(2)-4=32\\\\N'(6)=6(6)^2+6(6)-4=248

so the turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

The turtle population at the end of the tenth year will be

N(10)=2(10)^3+3(10)^2-4(10)+1000\\N(10)=3260 \:turtles

3 0
2 years ago
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