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2 years ago

BELP ME WITH THIS PLZ!!!!!!

Mathematics

1 answer:

kati45 [8]2 years ago

3 0

Answer:

The entire area of the sailboat is 60cm²

Step-by-step explanation:

You can find the area of this shape by breaking it down into simpler shapes and adding up their individual areas.

In this case, the areas we'll use are the rectangle at the bottom, and the pair of triangles at the top.

Because the two triangles can be put together to form a single triangle, we don't need to measure them independently. We can simply take the total length of their bases, multiply it by their height, and divide by two. This follows the rule that the area of a triangle is equal to the area of the square that contains it divided by two.

(2cm + 3cm) × 6cm

= 5cm × 6cm

= 30cm²

The rectangle's area is of course equal to its width times its height, so we can say:

2.5cm × 12cm

= 30cm²

The total area of the shapes then is 30cm² + 30 cm², giving us a total area of 60cm²

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Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago

What is the vertex of y<∣x−3∣+5

Tcecarenko [31]

Answer:

(3, 5)

Step-by-step explanation:

The graph is is the standard y=|x| except the values tells you that x shifts 3 (within the absolute value or parentheses x does the opposite) to the right and the y value shifts 5 up (numbers outside parentheses affects y and does what it says). You can try using a table of values then graphing to check your answer.

3 0

2 years ago

An electronic store sells a large flat screen television for $1,699. Last month, the store sold 8 of these television sets. Abou

serg [7]

1699 / 8 = 212.375 and then round $212.38 per TV

6 0

3 years ago

Read 2 more answers

Which trigonometric functions are negative in the fourth (IV) quadrant?

anzhelika [568]

<h3>Answers are: sine, tangent, cosecant, cotangent</h3>

Explanation:

On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.

Since sine is negative, so is cosecant as this is the reciprocal of sine

csc = 1/sin

In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.

Because tangent is negative, so is cotangent.

The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.

7 0

2 years ago

Find the domain and range of the function represented by the graph

lora16 [44]

Answer:

Step-by-step explanation:

(4). <em>(b)</em> Domain is {0, 1, 2, 3} and Range is { - 3, - 2, - 1, 0}

(5). <em>(a)</em> Domain is [ - 4, 2] and Range is [ - 4, 4]

domain: - 4 ≤ x ≤ 2 , range: - 4 ≤ y ≤ 4

6 0

2 years ago