What is x? 26.7% of x=12

it costs $20 per hour to bowl and $3 for shoe rental. Write a verbal model and an algebraic expression to represent the cost for

katen-ka-za [31]

3 + 20x, where x= the number of hours bowled. A verbal diagram is where you say amount to pay per hour * number of hours + price of shoe rental = total cost.

5 0

2 years ago

What is the answer to 3/4 +1/3

Elis [28]

Well, to add fractions, you need to find a common denominator. In this case, the smallest common denominator would be 12. So you must multiply each fraction so that both denominators are 12.
3/4*3/3=9/12
1/3*4/4=4/12
Add those two fractions together, reduce if possible, and you have your answer!
9/12+4/12=13/12
You can't reduce, so 13/12 is your answer.
Hope it helps!
-Lacy

3 0

3 years ago

HELPPP IF YOU CAN DO THISSSS (BRAINLIST) thank youuu

grigory [225]

Answer:

a) $50,880

b) $48

c) 1060

Step-by-step explanation:

a)

-320x^2 + 1920x + 48000 = 0

This is a parabola that opens downward. It has a maximum value. The maximum value occurs at x = -b/(2a)

x = -b/(2a) = (-1920)/(2(-320)) = 1920/640 = 3

The maximum revenue, y, is the value of the function evaluated at x = 3.

f(x) = -320x^2 + 1920x + 48000

f(-3) = -320(3)^2 + 1920(3) + 48000

f(-3) = 50,880

The maximum revenue is $50,880

b)

Since maximum revenue occurs at x = 3, and since x represents the number of $4 discounts, the discount is 3 * $4 = $12. The price is $60 - $12 = $48

c)

$50,880/$48 = 1060

3 0

3 years ago

What is the value of x geometry PLSS HELPP

Bogdan [553]

The interior angle measures are 900 degrees
131+122+125+148+107+126 = 759
900-759 =141
X = 141

7 0

2 years ago

3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared

navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

$\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end$

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

8 0

2 years ago