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4 years ago

Which of the statements below best describes direct variation?

Mathematics

1 answer:

REY [17]4 years ago

8 0

Direct variation is when one variable changes the other changes in proportion of the first, therefore all the above

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How do you write 144 by using 12 as the base

natima [27]

The answer would be <span> because you are using 12 as the base and 12 x 12 = 144 but hoped this helped :) :D :)</span>

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3 years ago

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Please anyoneeeeeeee.....i am almost done just please help me

VMariaS [17]

Answer:

B. XV and XY

Step-by-step explanation:

you can eliminate the top two options because those aren't rays. XV and XY are the only ones that end in X

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3 years ago

Jack’s bicycle tires have a diameter of 24 inches. If he rides at 15 miles per hour, what is the angular velocity of the wheels

Oksanka [162]

Answer:

Option 3 ⇒ 210.08 rpm

Step-by-step explanation:

The relation between the angular velocity ω and the linear velocity v is v=ωr

Where r is the radius of the tire.

Given that a diameter of 24 inches. If he rides at 15 miles per hour.

∴ r = diameter/2 = 24/2 = 12 in.

And v = 15 miles/hour

Converting the speed to inches per minutes where mile = 63,360 inches and hour = 60 minuted

∴ v = 15 * 63,360/60 = 15,840 inches/minute

∴ ω = v/r = 15,840/12 = 1,320 rad/minute

Converting ω from rad per minutes to revolutions per minute

Where 1 revolution = 2π

∴ ω = 1,320 / (2π) = 210.08 rpm

7 0

3 years ago

dos empresas de computadoras tienen el mismo precio en la lista de artículo una empresa ofrece El 25% y el 15% de descuento la o

Luden [163]

Answer:

Step-by-step explanation:

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3 years ago

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

6 0

4 years ago