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3 years ago

7

If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle,

accident investigators can estimate the speed of the car. To find the radius, accident investigators choose points A and B on the tire marks. Then, the investigators find the midpoint C of AB¯¯¯¯¯¯¯¯. Use the diagram to find the radius r of the circle. Round your answer to the nearest tenth.

1 answer:

Slav-nsk [51]3 years ago

6 0

Answer:

155.7

Step-by-step explanation:

Use what you know.

Segment AC is 130 ft

Segment CD is 70ft

If you use the Pythagorean Theorem, in this case being = +

To find segment CE, you would do r-70

So, = +

= 16,900 + -140r + 4900

Add the -140r to the left side and then get rid of the two . Then Add 16,900 and 4900 together

You'll end up with 140r = 21,800

Divide 140 on each side.

Your final answer will be 155.7 (rounded to the nearest tenth)

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3 years ago

Can someone help 6-11x=7x-12

Luden [163]

6-11x = 7x-12

6+12 = 11x+7x

18 = 18x

x = 1

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3 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

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2 years ago

Part B: Your cell phone and tablet use the same account. The cell phone costs $0.10 for

Oksanka [162]

The inequalities are x + y < 5000 and 0.10x + 0.15y ≤ $450 which shows each month, you want to only spend up to $450 and use less than 5,000 minutes on both of the devices.

<h3>What is inequality?</h3>

Inequality is defined as a mathematical expression in which both sides are not equal and have mathematical signs that are either less than or greater than.

Let's suppose the total minutes of uses for cell phone is x

and total minutes of uses of for tablet is y

Cell phone cost $0.10 each minute and tablet cost $0.15 each minute.

Then total minutes will be:

x + y < 5000 (as it is given that the time will be less than 5,000 minutes on both of the devices)

And total cost = $450

0.10x + 0.15y ≤ $450

Thus, the inequalities are x + y < 5000 and 0.10x + 0.15y ≤ $450 which shows each month, you want to only spend up to $450 and use less than 5,000 minutes on both of the devices.

Learn more about the inequality here:

brainly.com/question/19491153

#SPJ1

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3 years ago