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goblinko [34]
3 years ago
9

What ratio is equivalent to 9:6 ? 12:8 7:4 3:1 15:10

Mathematics
2 answers:
slega [8]3 years ago
7 0

Answer:

15:10

Step-by-step explanation:

15:10 can be reduced into 3:2 by dividing with 5

and 9:6 can be reduced to 3:2 by dividing with 3

keep smiling.

mina [271]3 years ago
7 0

Answer:

D

Step-by-step explanation:

just got in edg 2021

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I need help pleases I wil give brainyless
erica [24]

Answer: Let's start by writing down coordinates of all points:

A(0,0,0)

B(0,5,0)

C(3,5,0)

D(3,0,0)

E(3,0,4)

F(0,0,4)

G(0,5,4)

H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate  changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)

Reflecting

A(0,0,0)

B(0,5,0)

Reflecting

B(0,-5,0)

C(3,5,0)

Reflecting

C(3,-5,0)

D(3,0,0)

Reflecting

D(3,0,0)

b.)

A(0,0,0)

Moving

A(-2,-3,1)

B(0,-5,0)

Moving

B(-2,-8,1)

C(3,-5,0)

Moving

C(1,-8,1)

D(3,0,0)

Moving

D(1,-3,1)

Hope this helps plz mark brainliest

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find the value of X and y<br>☺ -3/2y = -13<br>6X2Y = 4<br>by elimination​
alina1380 [7]
X equals 13/3 and Y equals 1/13 sorry if I am wrong
4 0
3 years ago
Keri is making a doll clothes for a holiday craft show. The wholesale cost of the materials for one outfit is $9.38. If she sell
irinina [24]

Answer:

60%

Step-by-step explanation:

If I get you right, you're asking for the percentage gain to the nearest percentage, right?

Now to do this, we need to get the amount of profit she made. That's quite easy, we only need to subtract the cost price from the selling price. And that equals $15 - $9.38 that equals $5.62.

Now, we divide this value by the cost price and multiply by 100% to get the percentage gain.

I.e 5.62/9.38 × 100%.

That gives 59.9% , which equals 60% when approximated to the nearest percentage.

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
Why is there an AA similarity postulate but not an AA congruence postulate
bagirrra123 [75]

Answer:

Step-by-step explanation:

4 0
3 years ago
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