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3 years ago

The picture below shows a right-triangle-shaped charging stand for a gaming system:

Mathematics

1 answer:

Setler79 [48]3 years ago

3 0

This question is incomplete because it lacks the diagram of the right angles shaped charging stand.

Find attached to this answer, the appropriate diagram.

Answer:

3(tan 50°)

Step-by-step explanation:

To solve this question, we would be using the Trigonometric function called Tangent.

tan θ = opposite side/ adjacent side

θ = 50°

Opposite side (Height)= Side AC = Unknown

Adjacent side = Side BC = 3 inches

Hence,

tan 50° = Side AC/ Side BC

tan 50° = Side AC / 3

Cross multiply

tan 50° × 3 = Side AC

3(tan 50°) = Side AC

Therefore, the expression shows the height, AC, of the charging stand is

3(tan 50°).

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3 years ago

Need help! Law of cosines...find b round to the nearest tenth please

dusya [7]

ANSWER

$B = 27.8 \degree$

EXPLANATION

The law of cosine is given by the formula:

${b}^{2} = {a}^{2} + {c}^{2} - 2bc \cos(B)$

From the diagram,

a=17, b=8, c=16

We substitute the given values into the formula to obtain:

${8}^{2} = {17}^{2} + {16}^{2} - 2(17)(16)\cos(B)$

$64 = 289+ 256- 544\cos(B)$

$64 = 545- 544\cos(B)$

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3 0

3 years ago

Read 2 more answers

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

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

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