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3 years ago

Evaluate

Mathematics

1 answer:

VMariaS [17]3 years ago

3 0

Answer:

-3/4

Step-by-step explanation:

A calculator can work this for you. All you need to do is substitute the numbers for the variables.

yx^y -xy^x

= 4(1/2)^4 -(1/2)(4^(1/2))

= 4/16 - 1/2√4 = 1/4 - 1

= -3/4

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Solve the following triangle. Given A=51 degrees b=40 c=45

STALIN [3.7K]

Answer:

$a=36.87\ units$

$B=57.47^o$

$C=71.53^o$

Step-by-step explanation:

step 1

Find the length side a

Applying the law of cosines

$a^2=b^2+c^2-2(b)(c)cos(A)$

substitute the given values

$a^2=40^2+45^2-2(40)(45)cos(51^o)$

$a^2=1,359.4466$

$a=36.87\ units$

step 2

Find the measure of angle B

Applying the law of sines

$\frac{a}{sin(A)} =\frac{b}{sin(B)}$

substitute the given values

$\frac{36.87}{sin(51^o)} =\frac{40}{sin(B)}$

$sin(B)=\frac{sin(51^o)}{36.87}{40}$

$B=sin^{-1}(\frac{sin(51^o)}{36.87}{40})=57.47^o$

step 3

Find the measure of angle C

Remember that the sum of the interior angles in any triangle must be equal to 180 degrees

$A+B+C=180^o$

substitute the given values

$51^o+57.47^o+C=180^o$

$108.47^o+C=180^o$

$C=180^o-108.47^o=71.53^o$

5 0

3 years ago

X+2y=-1 and 4x-4y=20 solve by substitution

Illusion [34]

To do this problem you would first need to factor out a variable, which in this case I would want to do the first equation because it is isolated. Now the equations would look like this:

x = -2y - 1

4x - 4y = 20

Since we know that x is now equal to -2y - 1 we can plug it in to the x value in the second equation:

4 (-2y -1) - 4y = 20

-8y -4 - 4y

-12y - 4 = 20

-12y = 24

y = -2

Now that we know the y value plug the y value to one equation to find the x, I will be using the first equation

x + 2(-2) = -1

x - 4 = -1

x = 3

Solutions:

y = -2

x = 3

8 0

3 years ago

Pleaseeee helpp which one is it and whyyyy!!!!

Elza [17]

Answer:

answer is A

Step-by-step explanation:

7 0

3 years ago

Dominique wants to use the distributive property to mentally find the value of

GalinKa [24]

She can use

15(37+ 63)

Hope this helps!

5 0

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 $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

3 years ago