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3 years ago

What is the domain of the radical function shown on the graph? GRAPH IS IN THE PIC BELOW. A.

Mathematics

1 answer:

irina1246 [14]3 years ago

8 0

Given:

The graph of a radical function.

To find:

The domain of the given radical function.

Solution:

We know that, domain is the set of input values or we can say domain is the set of x-values for which the function is defined.

From the given graph it is clear that, for each value of x there is a y-value. It means the function is defined for all real values of x. So,

Domain = Set of all real numbers.

Therefore, the correct option is A.

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Write the complex number in polar form with argument between 0 and 2.

pychu [463]

Lets say that √(3+i) is the magnitude of your vector. In polar form, i represents the x component and j represents the y component of the vector. Therefore, the polar form is icosθ√(3+i) + jsinθ√(3+i)

draw a diagram :

z = r(cos θ + i sin θ), where z = complex number, r = modulus, θ = angle of rotation For the given:r = 2θ = π/6 √3 + i = 2(cos π/6 + i sin π/6)

hope thz hlpz ✿

5 0

2 years ago

What information do you need to compare the services of different banks

Firdavs [7]

You would need to know customer reviews, savings and checking accounts, fees, and CD rates in order to find the services of different banks.

8 0

3 years ago

On a town map, each unit of the coordinate plane represents 1 mile. Three branches of a bank are located at A(−3, 1), B(3, 3), a

Andrei [34K]

Answer:

The minimum total distance the employee may have driven before getting stuck in traffic is 8.3 miles

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of the distance between to points $(x_{1},y_{1})$

and $(x_{2},y_{2})$ i s $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

- So at first we will find the distance between the three points

∵ A = (-3 , 1) , B = (3 , 3) , C = (3 , -1)

- By using the rule of distance

∴ $AB=\sqrt{(3--3)^{2}+(3-1)^{2}}=\sqrt{36+4}=\sqrt{40}=6.325$

∴ $BC=\sqrt{(3-3)^{2}+(-1-3)^{2}}=\sqrt{0+16}=4$

∴ $AC=\sqrt{(3--3)^{2}+(-1-1)^{2}}=\sqrt{36+4}=\sqrt{40}=6.325$

∵ Each unit coordinate plane represents 1 mile

∵ AB = 6.3 units

∴ The distance between branches A and B = 6.325 miles

∵ BC = 4 units

∴ The distance between branches B and C = 4 miles

∵ AC = 6.3 units

∴ The distance between branches A and C = 6.325 miles

- A bank employee drives from Branch A to Branch B and then

drives halfway to Branch C before getting stuck in traffic

∵ The distance from branch A to branch B is 6.325 miles

∵ The distance from branch B to branch C is 4 miles

∵ The half way from branch B to branch C = 1/2 × 4 = 2 miles

∴ The distance the employee may have driven before getting

stuck in traffic = 6.325 + 2 = 8.325 miles

∴ The minimum total distance the employee may have driven

before getting stuck in traffic is 8.3 miles

8 0

3 years ago

A 1300 kg car moving at 5.2 m/s is initially traveling north in the positive y direction. After completing a 90° right-hand turn

stich3 [128]

Answer:

a. 6760(i - j) kgm/s b. -6760i kgm/s c. 2172.74 N d. 19314.29 N e. 45°

Step-by-step explanation:

Given that mass of car, m = 1300 kg

initial velocity in the y-direction, v₁ = 5.2j m/s

time taken for 90° turn to positive x-direction, t₁ = 4.4 s

time taken for collision in positive x-direction, t₂ = 350 ms

a. Impulse, J₁ on car during turn.

impulse, J =mv₂ - mv₁

v₁= initial velocity and v₂ = final velocity

v₁ = 0i + 5.2j m/s since it was initially moving in the positive y-direction.

v₂ = 5.2i + 0j m/s since it was initially moving in the positive x-direction.

So, J₁ = m(5.2i + 0j -(0i + 5.2j))

= m(5.2i - 5.2j)

= 1300× 5.2(i-j)

= 6760(i-j) kgm/s

b. Impulse after collision J₂

v₁= initial velocity=5.2i + 0j m/s since it was initially moving in the positive x-direction.

v₂= 0m/s since the car stops after collision

So, J₂= mv₂ - mv₁

= m(0 - (5.2i + 0j)) m/s

=-5.2mi kgm/s

= -1300 × 5.2i kgm/s

=-6760i kgm/s

c. Average force, F₁ during turn

Impulse J = Ft

From (a) the impulse J₁ = 6760(i-j) kgm/s. The time taken for the turn t₁ = 4.4 s. So, F₁ = J₁/t₁ = 6760(i-j)/4.4 = 1536.36(i-j) N.

Magnitude of F₁ = F₁ = average force during turn=1536.36√2= 2172.74 N

d. Average force, F₂ after collision

From J=Ft, F=J/t

From (b) above, our impulse during collision is J₂ = -6760i kgm/s. The time taken for the impulse or collision to occur is t₂ = 350 ms.

So, F₂ = J₂/t₂ = -6760i /(350 × 10⁻³) N= - 19314.29i N.

So magnitude of F₂= F₂=average force during collision = 19314.29 N

e. Angle between average force in (c) and the positive x- direction.

We know that F₁= 1536.36(i-j) N = 1536.36i - 1536.36j N. The unit vector in the positive x-direction is i.

For the angle between two vectors, we have that cosθ= a.b/ab where a,b are vectors and a,b their magnitudes respectively.

So, cosθ = (1536.36i - 1536.36j).i/(1536.36√2) =1536.36/1536.36√2=1/√2

cosθ = 1/√2

θ=cos⁻¹(1/√2)

θ=45°

4 0

3 years ago

Wanda starts with $1200 and deposits $40 a week. Yolanda starts with $500 and deposits $60 per week. When will Wanda have the sa

boyakko [2]

Week 35.

They will both have $2600 dollars by week 35.

Hope this helps!

3 0

3 years ago