Find the value of the rectangular prism

Which table shows a proportional relationship?? help asap

larisa [96]

Answer:

i wanna say its the first one if its wrong im very sorry

Step-by-step explanation:

5 0

3 years ago

2.

abruzzese [7]

Answer:

a) 30 kangaroos in 2030

b) decreasing 8% per year

c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo

Step-by-step explanation:

We assume your equation is supposed to be ...

P(t) = 76(0.92^t)

__

a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30

In the year 2030, the population of kangaroos in the province is modeled to be 30.

__

b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.

The population is decreasing by 8% each year.

__

c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.

P(100) = 75(0.92^100) = 76(0.0002392)

P(100) ≈ 0.0182, about 1/55th of a kangaroo

5 0

3 years ago

Match the following rational expressions to their rewritten forms.

Nookie1986 [14]

Answer:

Answer image is attached.

Step-by-step explanation:

Given rational expressions:

$1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}$

And the rewritten forms:

$(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}$

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

$(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}$

So, correct option is 3.

$(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}$

So, correct option is 1.

$(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}$

So, correct option is 2.

$(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}$

So, correct option is 4.

The answer is also attached in the answer area.

7 0

2 years ago

Aanya lives in Ohio. She immigrated with her family from India to North America. Aanya wants to visit a cousin in Toronto by dri

Genrish500 [490]

Answer:

(b) 310 mi, 12.76km/liter

Step-by-step explanation:

We are told that Toronto is 500 km north on Highways. Since 1 km = 0.62 mi, then Toronto is: 500 × 0.62 = 310 miles

Aanya’s car averages 30 mi/gallon.

We are told that 1 mi = 1.61 km

Thus, her average is now;

30 × 1.61 km/gallon = 48.3 km/gallon

We are told that 1 qt = 0.946 liters

But also, 0.25 gallon = 1 qt

Thus;

0.25 gallon = 0.946 litres

Thus, 1 gallon = 0.946/0.25 litres = 3.784 litres

Thus, 48.3 km/gallon to km/litres = 48.3/3.784 = 12.76 km/litres

Answer is 310 miles and 12.76 km/litres

7 0

3 years ago

Which of these values cannot represent the probability of an event happening?

ololo11 [35]

Answer:

D.

Step-by-step explanation:

The last option, 1.2, cannot represent the probability of an event happening since it is more than 100% (120%). And event cannot have a higher probability of happening than "happening".

A. 0.33 is a valid response, since it represents 33%.

B. 56% is also a valid response.

C. 7/8 is also a valid response since it represents 0,875 or 87,5%.

Hope it helped,

BiologiaMagister

6 0

2 years ago

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