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3 years ago

According to this diagram, what is tan 74

Mathematics

2 answers:

tresset_1 [31]3 years ago

8 0

Answer: F-24/7

Step-by-step explanation: sahcohtoa

tan is opposite over adjacent so 24/7

mariarad [96]3 years ago

7 0

Answer:

$\boxed{F.\:\:\frac{24}{7} }$

Step-by-step explanation:

The tangent ratio is the ratio of the length of the opposite side to the length of the adjacent side.

This implies that;

$\tan(74\degree)=\frac{Opposite}{Adjacent}$

The length of the side opposite to the 74 degree angle is 24 units.

The length of the side adjacent to the 74 degree angle is 7 units.

We substitute these values into the formula to obtain;

$\tan(74\degree)=\frac{24}{7}$

The correct answer is option F

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Answer:

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52 + 4a = 62

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2 years ago

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3 years ago

An account with an initial balance of $5000 earns 5% interest for 3.5 years.

goldenfox [79]

Answer:

The investor will earn $875 over the course of 3.5 years.

Step-by-step explanation:

Sounds like simple interest (not compound interest).

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3 years ago

Read 2 more answers

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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3 years ago

What is -32 divided by 9

soldier1979 [14.2K]

The answer is −3.555555555555556 but you can just put -3.55

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3 years ago

Read 2 more answers