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anastassius [24]

3 years ago

9

A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 87 m long and 69 m wide.

What is the length of a training track running around the field? (Use the value 3.14 for π, and do not round your answer. Be sure to include the correct unit in your answer.)

1 answer:

professor190 [17]3 years ago

7 0

Answer:

607.32

Step-by-step explanation:

two semi circles make a whole circle so use the normal circumfrence formula 2 pi r

circumfrence = 433.32

the perimeter of the rectangle is 87 + 87, because they are on the outside of the field.

The answer is 607.32

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The area of a piece of land that is in the shape of a triangle is 1/12 square mile. One dimension of this piec of land is 4/21.

Arisa [49]

Answer:

D2=21/24 mile

Step-by-step explanation:

Her it is not specified if it is base or height, it only talks about dimensions, so let's work with dimensions

A=1/12 $mile^{2}$

D1=4/21mile, D2=?

Area of a triangle $A=\frac{b.h}{2}$

A=((D1*D2)/2) → 2A=D1*D2 → D2=2A/D1 so D2=(2*1/12)/(4/21) =(1/6)/(4/21) = (1*21)/(6*4)=21/24 mile

7 0

3 years ago

0.2(x+50)-6=0.4(3x+20)<br><br> ( show work please)

garri49 [273]

Let's solve your equation step-by-step.<span><span><span>0.2<span>(<span>x+50</span>)</span></span>−6</span>=<span>0.4<span>(<span><span>3x</span>+20</span>)</span></span></span>Step 1: Simplify both sides of the equation.<span><span><span>0.2<span>(<span>x+50</span>)</span></span>−6</span>=<span>0.4<span>(<span><span>3x</span>+20</span>)</span></span></span><span><span><span><span><span><span>(0.2)</span><span>(x)</span></span>+<span><span>(0.2)</span><span>(50)</span></span></span>+</span>−6</span>=<span><span><span>(0.4)</span><span>(<span>3x</span>)</span></span>+<span><span>(0.4)</span><span>(20)</span></span></span></span>(Distribute)<span><span><span><span><span>0.2x</span>+10</span>+</span>−6</span>=<span><span>1.2x</span>+8</span></span><span><span><span>(<span>0.2x</span>)</span>+<span>(<span>10+<span>−6</span></span>)</span></span>=<span><span>1.2x</span>+8</span></span>(Combine Like Terms)<span><span><span>0.2x</span>+4</span>=<span><span>1.2x</span>+8</span></span><span><span><span>0.2x</span>+4</span>=<span><span>1.2x</span>+8</span></span>Step 2: Subtract 1.2x from both sides.<span><span><span><span>0.2x</span>+4</span>−<span>1.2x</span></span>=<span><span><span>1.2x</span>+8</span>−<span>1.2x</span></span></span><span><span><span>−<span>1x</span></span>+4</span>=8</span>Step 3: Subtract 4 from both sides.<span><span><span><span>−<span>1x</span></span>+4</span>−4</span>=<span>8−4</span></span><span><span>−<span>1x</span></span>=4</span>Step 4: Divide both sides by -1.<span><span><span>−<span>1x</span></span><span>−1</span></span>=<span>4<span>−1</span></span></span><span>x=<span>−4</span></span>Answer:<span>x=<span>−<span>4</span></span></span>

7 0

3 years ago

Read 2 more answers

I have 273 points, I ask a question that only 2 people can answer, if the question is worth 100 points, how much would each pers

Mariulka [41]

I think 171 but I’m not sure

6 0

2 years ago

Read 2 more answers

A bridge in the shape of an arch connects two cities separated by a river. The two ends of the bridge are located at (–7, –13) a

sdas [7]

Answer:

$y=-\dfrac{13}{49}x^2$

Step-by-step explanation:

The shape of an arch corresponds to a parabola.

the general equation for a parabola is:

$y=ax^2+bx+c$

we're given three coordinates: (-7,-13),(7,-13) and (0,0)

so we can plug these values in the general equation to make 3 separate equations:

(x,y) = (-7,-13)

$-13=a(-7)^2+b(-7)+c$

$49a-7b+c=-13$

(x,y) = (7,-13)

$-13=a(7)^2+b(7)+c$

$49a+7b+c=-13$

(x,y) = (0,0)

$0=a(0)^2+b(7)+c$

$c=0$

so we have three equations. and we can solve them simultaneously to find the values of a,b, and c.

we've already found c = 0, let's use substitute it to other equations.

$49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13$

$49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13$

we can solve these two equation using the elimination method, by simply adding the two equations

$\quad\quad49a-7b=-13\\+\quad49a+7b=-13$

------------------------------

$\quad\quad 98a=-26$

$\quad\quad a=-\dfrac{13}{49}$

Now we can plug this value of a in any of the two equations.

$49a-7b=-13$

$49\left(-\dfrac{13}{49}\right)-7b=-13$

$-13-7b=-13$

$-7b=0$

$b=0$

We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.

$y=\left(-\dfrac{13}{49}\right)x^2+0x+0$

$y=-\dfrac{13}{49}x^2$

$49y=-13x^2$

This our equation of the arch!

5 0

2 years ago

Can you please help

Dmitry [639]

Answer is in the file below

tinyurl.com/wpazsebu

4 0

2 years ago