Quick! Need help, anyone know this?

Evaluate the function.<br><br> g(x)=10x+3<br><br> g(19.6)=

Zolol [24]

Answer:

199

Step-by-step explanation:

plug in x=19.6, then

g(19.6) = 10 * 19.6 +3= 199

7 0

3 years ago

A rectangle has a height of 3 and a width of 2x*2 + 3x - 5 , express area of whole rectangle, please answer this ASAP !!!

monitta

Area is height time width, or in this case, 3(2x^2+3x+5), which is 6x^2+9x+15.

8 0

3 years ago

Someone please help me out:(

gogolik [260]

SAS mmmmmmmmmmmmmmmmmmm

5 0

3 years ago

Let f(x) = log(x). Find values of a such that f(kaa) = kf(a).

meriva

Answer:

a = $k^{\frac{1}{k-2}}$

Step-by-step explanation:

Given:

f(x) = log(x)

and,

f(kaa) = kf(a)

now applying the given function, we get

⇒ log(kaa) = k × log(a)

or

⇒ log(ka²) = k × log(a)

Now, we know the property of the log function that

log(AB) = log(A) + log(B)

and,

log(Aᵇ) = b × log(A)

Thus,

⇒ log(k) + log(a²) = k × log(a) (using log(AB) = log(A) + log(B) )

or

⇒ log(k) + 2log(a) = k × log(a) (using log(Aᵇ) = b × log(A) )

or

⇒ k × log(a) - 2log(a) = log(k)

or

⇒ log(a) × (k - 2) = log(k)

or

⇒ log(a) = (k - 2)⁻¹ × log(k)

or

⇒ log(a) = $\log(k^{\frac{1}{k-2}})$ (using log(Aᵇ) = b × log(A) )

taking anti-log both sides

⇒ a = $k^{\frac{1}{k-2}}$

3 0

3 years ago

Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona

andrey2020 [161]

Answer:

a. v.v' = v₁v₂ - v₁v₂ = 0 b. (20, -21)/29 and (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' = (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

= v₁v₂ - v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0

2 years ago