Ask question

Ask question

All categories

2 years ago

8

Could someone pls help me thankss

1 answer:

sergey [27]2 years ago

7 0

$792x .06= 47.52
the total price is $839.52

You might be interested in

the population of center city is modeled by exponential function f, where x is the number of years after the year of 2015. the g

lora16 [44]

Answer:

the Answer is B) y >= to 250,000

5 0

2 years ago

Me. Silvera bought a can of paint. He used 3/8 of it to paint a fence and 1/4 to paint a table top. He painted a chair with some

LuckyWell [14K]

Well, seeing the information above, he used 5/8 of paint to paint the fence and table top. If he only had 1/8 left of the paint, he used 2/8 on the chair.

I found the 5/8 because I needed a common denominator. I multiplied the top and bottom by 2 to get 2/8. 3/8+2/8=5/8, and then I added 2/8 to the 5/8 that then gave me 7/8 of paint used, and 1/8 of paint left.

Hope I helped!

3 0

3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

2 years ago

For 0 ≤ θ < 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)

gregori [183]

Recall the half-angle identity for sine,

sin²(x/2) = (1 - cos(x))/2

Then the given equation is identical to

sin²(θ) = 1 - cos(θ)

Also recall the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Then we rewrite the equation as

1 - cos²(θ) = 1 - cos(θ)

Factoring the left side, we have

(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)

and so

(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0

and we factor this further as

(1 - cos(θ)) (1 + cos(θ) - 1) = 0

which gives

cos(θ) (1 - cos(θ)) = 0

Then either

cos(θ) = 0 or 1 - cos(θ) = 0

cos(θ) = 0 or cos(θ) = 1

[θ = arccos(0) + 2nπ or θ = -arccos(0) + 2nπ]

… or [θ = arccos(1) + 2nπ or θ = -arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ or θ = -π/2 + 2nπ] or [θ = 0 + 2nπ]

In the interval 0 ≤ θ < 2π, we get three solutions:

• first solution set with n = 0 ⇒ θ = π/2

• second solution set with n = 1 ⇒ θ = 3π/2

• third solution set with n = 0 ⇒ θ = 0

So, the first choice is correct.

6 0

2 years ago

Compare the value of 495,123 and 63,129

Vesnalui [34]

63,219 is overall the lower value, but has the greater hundreds and ones place.

Hope this helps!

8 0

3 years ago