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8090 [49]
3 years ago
11

Which statement is true? All quadrilaterals have at least 1 pair of parallel sides. Parallelograms have 2 pairs of congruent ang

les. A rhombus is a rectangle. A trapezoid is a parallelogram.
Mathematics
2 answers:
quester [9]3 years ago
6 0

Answer:

all quadrilaterals have at least one pair of parallel sides

Step-by-step explanation:

this this is true because all quadrilaterals starting from triangle has at least one pair of parallel sides

QveST [7]3 years ago
6 0

Answer:

PARALLELOGRAM have 2 pairs of congruent angles

Step-by-step explanation:

QUADRILATERALS have one pair of parallel lines. If it has two sets, it NOW become a PARALLELOGRAM.

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Trapezoid ABCD is congruent to trapezoid A"B"C"D". Which sequence of transformations could have been used to transform trapezoid
Contact [7]

Answer:ygggydtdgdgdfddgfs

Step-by-step explanation:

8 0
2 years ago
2 - 2n = 3n + 17 solve n
Masja [62]

Answer:

n = -3

Step-by-step explanation:

2 - 2n = 3n + 17

-2                -2

-2n = 3n + 15

-3n    -3n

<u>-5n = 15</u>

-5       -5

n = -3


Hope this helps you!!! :)

4 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
(LC)
serg [7]
10

1240 x 10 = 12400
12400 = 12400
8 0
3 years ago
Solve the proportion
sergeinik [125]

y =  \frac{28}{5}

4 0
2 years ago
Read 2 more answers
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