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kipiarov [429]
3 years ago
6

Which pair of undefined terms is used to define the term parallel lines?

Mathematics
1 answer:
Nikolay [14]3 years ago
5 0
<span>After a thorough research, there exists the same question that have choices.

</span><span>Point, line
Line, plane
Coplanar, line
Point, coplanar </span><span>

The correct answer is "Coplanar, Line." The pair of undefined terms that is used to define the term parallel line is "Coplanar, Line" because parallel lines are coplanar in itself that do not intersect.
</span><span>

</span>
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In Texas, the population of a species of bats is decreasing at a rate of -0.021 per year. The population was 22,000 in 2010. Acc
Shtirlitz [24]

Answer:

19,807

Step-by-step explanation:

4 0
3 years ago
HELP WITH ANY OR ALL THESE PROBLEMS ASAP PLEASE LIKE NOW PLEASE<br>CAN SOMEONE PLEASE HELP 
LiRa [457]

Ques 8:

The Volume(V) in cubic feet of an aquarium id modeled by the polynomial function V(x)= x^{3}+2x^{2}-13x+10

a) We have to explain that why x =4 is not a possible rational zero.

By Factor theorem, which states that a polynomial f(x) has a factor (x - k) if and only if f(k)=0.

For this , we will substitute the value of x in the given function.

V(x)=x^{3}+2x^{2}-13x+10

V(4)=4^{3}+2(4)^{2}-13(4)+10

V(4)=4^{3}+2(4)^{2}-13(4)+10

V(4)=54 which is not equal to zero.

Therefore, x=4 is not a possible rational zero.

(b) To show that (x-1) is a factor of V(x).

By Factor theorem, which states that a polynomial f(x) has a factor (x - k) if and only if f(k)=0.

Let (x-1)=0

So, x=1.

Substituting x=1 in the given function.

V(1)=1^{3}+2(1)^{2}-13(1)+10

V(1)= -10+10

V(1) = 0

Therefore, (x-1) is a factor of V(x).

Now we will factorize the given function.

Dividing the given function by (x-1).

On dividing, we get quotient as x^{2}+3x-10

So, factored form is = (x-1)(x^{2}+3x-10)

= (x-1)(x^{2}+5x-2x-10)

= (x-1)(x(x+5)-2(x+5))

=(x-1)(x+5)(x-2)

(c) So, the dimensions are 1,2 and -5.

5 0
3 years ago
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mojhsa [17]
I- please ask a question. This is not a question
7 0
3 years ago
Read 2 more answers
Given that the mass of an average linebacker at Ursinus College is 250 lbs and the radius of a pea is 0.50 cm, calculate the num
Brilliant_brown [7]

Answer:

1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle

Step-by-step explanation:

Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is

V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³

the mass in that volume would be  m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)

The density of an alpha particle is  ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be

ρ= m/V

since both should be equal ρ=ρa , then

ρa= m/V =N*L/V → N =ρa*V/L

replacing values

N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg  = 1.544*10⁹ Linebackers

N=1.544*10⁹ Linebackers

7 0
3 years ago
What is the area to this shape ?
Dafna1 [17]

Answer:

249.42 units²

Step-by-step explanation:

Given only the apothem of an n-sided regular polygon, the area can be computed as ...

... A = n·a²·tan(180°/n)

For n=3 and a=4√3, this is ...

... A = 3·(4√3)²·tan(60°)

... A = 3·48·√3 = 144√3 . . . . units²

... A ≈ 249.42 units²

4 0
3 years ago
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