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2 years ago

Please help soon! I have a trig question...

Mathematics

1 answer:

larisa86 [58]2 years ago

5 0

The value of x will be 33 units approximately. Thus, option A is correct.

<h3>What are trigonometric identities?</h3>

Trigonometric identities are the functions that include trigonometric functions such as sine, cosine, tangents, secant, and, cot.

We need to find the value of x

so first

Tan 60 = Perpendicular / 11√6

Perpendicular = 1.732 (11√6)

Perpendicular = 46.66

Then,

Sin 45 = Perpendicular / x

Sin 45 = 46.66/ x

x = 33.00

So, the value of x will be 33 units approximately.

Thus, option A is correct.

Learn more about trigonometric;

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What is the value of P(B')?

Butoxors [25]

Answer: ) Events A and B are such that: P(A) = 0.4; P(B) = 0.5; P(A n B) = 0.3; and P(A u B) = 0.6. i) What is the value of P(A u B')? (Where B' is the compliment of B.) 0.

Step-by-step explanation: I just did a little extra credit :D

6 0

3 years ago

Use scalar multiplication to determine the coordinates of the vertices of the dilated figure. Then graph the pre-image and the i

Luba_88 [7]

Answer:

The coordinates of the vertices of the dilated figure are:

A' is (-2 , 4), B' is (4 , 8), C' is (4 , -2), D' is (-2 , -6) ⇒ the answer is (d)

Step-by-step explanation:

* Lets study the matrix of the dilation

- If we dilate any point by scale factor k we multiply the

coordinates of the point by k

- The matrix of the dilation by scale factor k is

$\left[\begin{array}{ccc}k&0\\0&k\end{array}\right]$

* Now lets solve the problem

- We will multiply the matrix of dilation by the matrix of the

vertices of the quadrilateral

- The dimension of the matrix of the dilation is 2×2 and the

dimension of the matrix of the vertices of the quadrilateral

is 2×4 then the dimension of the matrix of the image of the

quadrilateral is 2×4

∵ The scale factor is 2

∴ The matrix of dilation is $\left[\begin{array}{cc}2&0\\0&2\end{array}\right]$

∵ The matrix of the vertices of the quadrilateral is

$\left[\begin{array}{cccc}-1&2&2&-1\\2&4&-1&-3\end{array}\right]$

∴ The image of the quadrilateral is :

$\left[\begin{array}{cc}2&0\\0&2\end{array}\right]\left[\begin{array}{cccc}-1&2&2&-1\\2&4&-1&-3\end{array}\right]=$

$\left[\begin{array}{cccc}(2)(-1)+(0)(2)&(2)(2)+(0)(4)&(2)(2)+(0)(-1)&(2)(-1)+(0)(-3)\\(0)(-1)+(2)(2)&(0)(2)+(2)(4)&(0)(2)+(2)(-1)&(0)(-1)+(2)(-3)\end{array}\right]=$