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Kazeer [188]
3 years ago
8

Use the order of operations to simplify 3/4 + 8(2.5 - 0.5)

Mathematics
2 answers:
inna [77]3 years ago
5 0

Answer:

The correct answer is B

Step-by-step explanation:

3/4+8(2.5-0.5)

8(2.5-0.5)=16

=3/4-1+16

16 x 4+3=67

67/4=16 3/4

TiliK225 [7]3 years ago
3 0
The answer is B 16 3/4
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Answer:

3. [1, −2]

2. [−3, 3]

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Step-by-step explanation:

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{7⁄2x - ½y = 9⁄2

{3x - y = 5

-6⁄7[7⁄2x - ½y = 9⁄2]

{−3x + 3⁄7y = −3 6⁄7 >> New Equation

{3x - y = 5

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-4⁄7y = 1 1⁄7

-2 = y [Plug this back into both equations above to get the x-coordinate of 1]; 1 = x

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2.

{−3x + 9y = 36

{4x + 12y = 24

¾[4x + 12y = 24]

{−3x + 9y = 36

{3x + 9y = 18

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18y = 54

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18 18

y = 3 [Plug this back into both equations above to get the x-coordinate of −3]; −3 = x

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1.

{4x − y = −38

{x + y = 3

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x = -7 [Plug this back into both equations above to get the y-coordinate of 10]; 10 = y

I am joyous to assist you anytime.

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3 years ago
A car manufacturer rolled out a new car priced at $10,000, but not many people bought it. In the context of supply and demand, h
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Represent the following sentence as an algebraic expression, where "a number" is the letter x. You do not need to simplify.
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<h3>Answer:  (2x)^3</h3>

Work Shown:

x = some unknown number

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2 years ago
How to find the x and y intercepts for linear equations
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6 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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