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mafiozo [28]
3 years ago
14

Starting on Day 1 with 1 jumping jack, Kiera doubles the number of jumping jacks she does every day. How many jumping jacks will

Kiera do on Day 10?
Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

The answer is 512

Day 1:1

Day 2:2

Day 3:4

Day 4:8

Day 5:16

Day 6:32

Day 7:64

Day 8:128

Day 9:256

Day 10:512

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3(5z - 7) - 2 (9z - 11) = 4 (8z - 13) - 17
marysya [2.9K]

Answer z=2

See picture attached for the math

3 0
1 year ago
What is the circumference of a 45 m circle
Kobotan [32]
<span>to solve circumference of a circle is diameter times pi.
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3 years ago
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The vertices of a triangle are A(0, 0), B(3, 8), and C(9, 0). What is the area of this triangle?
Harlamova29_29 [7]
Answer: 36 units squared.


Explanation:
Area of a traingle is given by,

| 0.5{x1(y2-y3) + x2(y3-y1) + x3(y1-y2)} |

Accepting the values of the coordinates in x's and y's,

x1 =0, y1=0

x2=3, y2=8

x3 =9, y3=0

= | 0.5 × (-72)|
= 36 unit squared.
8 0
3 years ago
Read 2 more answers
3. There are 24 girls and 36 boys in Yuki's
Natasha2012 [34]

Answer:

a) 24:36 simplified: 2:3

b) 40%

c) 1%

Step-by-step explanation:

b) total people:

24 + 36= 60

24/60 = 0.4 40%

c) 1/60 = 0.01

01%

5 0
3 years ago
Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut
frozen [14]

Answer:

a)\sqrt{144-288t+208t^2} b.) -12knots, 8 knots c) No e)4\sqrt{13}

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved 12\dot t (n.m) and ship B has moved 8\dot t (n.m). We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}

b)

We want to find \frac{ds}{dt} for t=0 and t=1

\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0

Since function f(x)=199-288x+208x^2 is quadratic, concave up and has no real roots, we know that 199-288x+208x^2>0 for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

                                                \lim_{t \to \infty} \frac{ds}{dt}

So, lets check this limit:

\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}

Notice that:

4\sqrt{13}=\sqrt{12^2+5^2}=√(speed of ship A² + speed of ship B²)

5 0
3 years ago
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