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Reil [10]
3 years ago
12

I really need this ASAP

Mathematics
1 answer:
dmitriy555 [2]3 years ago
7 0

Answer:

x,y

(1,3)

Step-by-step explanation:

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Point T is at (-3, 8). What are the coordinates of T' after R(y-axis) o R(x-axis)?
Whitepunk [10]

Given:

Point is T(-3,8).

To find:

The coordinates of T' after R(y-axis)\circ R(x-axis).

Solution:

We know that, R(y-axis)\circ R(x-axis) means the figure reflected across the x-axis then reflected across y-axis.

If a figure reflected across x-axis, then

(x,y)\to (x,-y)

T(-3,8)\to T_1(-3,-8)

If a figure reflected across y-axis, then

(x,y)\to (-x,y)

T_1(-3,-8)\to T'(-(-3),-8)

T_1(-3,-8)\to T'(3,-8)

Therefore, the required point is T'(3,-8).

4 0
3 years ago
If 3x - 8= -2, find the value of x - 6.
siniylev [52]

Answer:

x-6 = -4

Step-by-step explanation:

3x - 8= -2

Add 8 to each side

3x - 8+8= -2+8

3x = 6

Divide by 3

3x/3 = 6/2

x =2

We want to find x-6

2-6 = -4

4 0
3 years ago
Read 2 more answers
Click on the solution set below until the correct one is displayed.
Snowcat [4.5K]

Answer: Second Option

(Point in Quadrant I)

Step-by-step explanation:

The solution to a system of linear equations is the point where the two lines intersect.

Note that in this case we have two lines with different slope . By definition, if two lines have different slopes and are contained in the same plane, then there will always be an intersection between them at some point in the plane.

Looking at the image, you can see that the lines get closer as x and y increase. Then they will intercept in the first quadrant.

8 0
3 years ago
Determine if each of the following sets is a subspace of Pn, for an appropriate value of n.
snow_tiger [21]

Answer:

1) W₁ is a subspace of Pₙ (R)

2) W₂ is not a subspace of Pₙ (R)

4) W₃ is a subspace of Pₙ (R)

Step-by-step explanation:

Given that;

1.Let W₁ be the set of all polynomials of the form p(t) = at², where a is in R

2.Let W₂ be the set of all polynomials of the form p(t) = t² + a, where a is in R

3.Let W₃ be the set of all polynomials of the form p(t) = at² + at, where a is in R

so

1)

let W₁ = { at² ║ a∈ R }

let ∝ = a₁t² and β = a₂t²  ∈W₁

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t²) + c₂(a₂t²)

= c₁a₁t² + c²a₂t²

= (c₁a₁ + c²a₂)t² ∈ W₁

Therefore c₁∝ + c₂β ∈ W₁ for all ∝, β ∈ W₁  and scalars c₁, c₂

Thus, W₁ is a subspace of Pₙ (R)

2)

let W₂ = { t² + a ║ a∈ R }

the zero polynomial 0t² + 0 ∉ W₂

because the coefficient of t² is 0 but not 1

Thus W₂ is not a subspace of Pₙ (R)

3)

let W₃ = { at² + a ║ a∈ R }

let ∝ = a₁t² +a₁t  and β = a₂t² + a₂t ∈ W₃

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t² +a₁t) + c₂(a₂t² + a₂t)

= c₁a₁t² +c₁a₁t + c₂a₂t² + c₂a₂t

= (c₁a₁ +c₂a₂)t² + (c₁a₁t + c₂a₂)t ∈ W₃

Therefore c₁∝ + c₂β ∈ W₃ for all ∝, β ∈ W₃ and scalars c₁, c₂

Thus, W₃ is a subspace of Pₙ (R)

8 0
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3 years ago
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