solve quadratic equation by factoring How do you do it 3k^2 + 72 = 33k

Answers

Answer 1

Answer: its going to be 33k -9k= 24k divided by 72 is 3. u do 72 ÷ 24 = 3

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Can someone help me on this im confused.. -3x-10=32

What is the value of 12.5 10^7?a. 12.5000000
b. 1,250,000
c. 125,000,000
d. 1.2500000

Answers

12.5 x 10^7 = 125,000,000
So the answer is C.)

love, the Pineapple.

A spherical balloon is inflating with helium at a rate of 6464piπ StartFraction ft cubed Over min EndFraction ft3 min. How fast is the balloon's radius increasing at the instant the radius is 22 ft? How fast is the surface area increasing?

Answers

Answer:

the surface area will increase 16 feet per minute

Step-by-step explanation:

Allow me to revise your question for a better understanding.

A spherical balloon is inflating with helium at a rate of 64π Start Fraction ft cubed Over min End Fraction ft3 min. How fast is the balloon's radius increasing at the instant the radius is 2 ft? How fast is the surface area increasing?

My answer:

Given that:

$(dV)/(dt) = 64\pi \frac{\text{ ft}^3}{\text{min}}$ (1)

As we know that, the volume of a spherical balloon can be calculated by using the following formula:

$V = (4)/(3)\pi r^3$ where r is the radius

Substitute V into (1), we have:

$(dV)/(dt) = (d)/(dt)((4)/(3)\pi r^3)\n\n(dV)/(dt) =4\pi r^2(dr)/(dt)$

In this situation, the instant radius is: 2f

So we have:

$64\pi = 4\pi (2)^2(dr)/(dt)\n\n\Rightarrow (dr)/(dt) = 4$

The radius of the balloon is increasing at a rate of 4 feet per minute

The the surface area has the formula as:

SA1 = 4π $r^(2)$

If the radius of the balloon is increasing at a rate of 4 feet per minute, so the

the surface area will increase 16 feet per minute because:

SA2 : 4π $(4r)^(2)$ = 16 (4π $r^(2)$ ) = 16 SA1

Hope it will find you well.

76 is what percent of 80

Answers

76 is 95% of 80, hope this helps you!

If cos θ -sin θ =1, find θ cos θ +sin θ?

Answers

$cos\theta-sin\theta=1\ \ \ \ |square\ both\ sides\n\n(cos\theta-sin\theta)^2=1\n\ncos^2\theta-2sin\theta cos\theta+sin^2\theta=1\n\n1-2sin\theta cos\theta=1\n\n-2sin\theta cos\theta=1-1\n\n-2sin\theta cos\theta=0\iff sin\theta=0\ or\ cos\theta=0\n\n\theta=k\pi\ or\ \theta=(\pi)/(2)+k\pi\ \ \ \ (k\in\mathbb{Z})$

$For\ \theta=k\pi\to cosk\pi-sink\pi=\pm1-0=\pm1\to conclusion:\theta=2k\pi\n\nFor\ \theta=(\pi)/(2)+k\pi\to cos((\pi)/(2)+k\pi)-sin((\pi)/(2)+k\pi)=0-(\pm1)=\mp1\to\n\to conclusion:\theta=(3\pi)/(2)+2k\pi\n\n============================\n\ncos\theta+sin\theta=(*)\n\nfor\ \theta=2k\pi\to(*)=1+0=1\n\nfor\ \theta=(3\pi)/(2)+2k\pi\to(*)=0-1=-1$

Does anyone know what the equation would be