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earnstyle [38]
2 years ago
5

Select the correct answer.

Mathematics
1 answer:
In-s [12.5K]2 years ago
5 0

Answer:

5-46i

Step-by-step explanation:

first multiply your (10-4i)(4-5i) so that would be 20-66i when simplified. then add that to -15+20i so it would look like 20-66i+(-15+20i). then simplify and you should get 5-46i.

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SNOG HELP OR SOMEONE THANK YOUUUU
dybincka [34]

Answer:

last one

Step-by-step explanation:

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32/2 = 16 miles per gallon WRONG

3/42 = 1/14 gallons per mile WRONG

42/3 = 14 miles per gallon RIGHT

7 0
3 years ago
Read 2 more answers
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
What are the coordinates of the vertex of the function below write your answer in the form y-9= -6(x-1)^2
Oliga [24]

Answer:

The co-ordinates of the vertex of the function y-9= -6(x-1)^2 is (1, 9)

<u>Solution:</u>

Given, equation is y-9=-6(x-1)^{2}

We have to find the vertex of the given equation.

When we observe the equation, it is a parabolic equation,

We know that, general form of a parabolic equation is  y-9=-6(x-1)^{2}

Where, h and k are x, y co ordinates of the vertex of the parabola.

\text { Now, parabola equation is } y-9=-6(x-1)^{2} \rightarrow y=-6(x-1)^{2}+9

By comparing the above equation with general form of the parabola, we can conclude that,

a = -6, h = 1 and k = 9

Hence, the vertex of the parabola is (1, 9).

7 0
3 years ago
S= 2m +2u <br><br> solve for m
Whitepunk [10]
M =0 if you put it in y= and hit 2nd graph and go down to the 0 there is a 1 with means m =0  (ti-84+ caculater)  
6 0
3 years ago
For the matrix a, find (if possible) a nonsingular matrix p such that p−1ap is diagonal.
Flura [38]
Sorry but can you please show like a diagram or something for me to answer it for you. Thank You

4 0
3 years ago
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