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Dimas [21]
2 years ago
9

identify the equation in the point-slope form the line parallel to y=3/4x-4 that passes through (-1,7)

Mathematics
1 answer:
hram777 [196]2 years ago
4 0
Y = 3/4x -4
The slope is 3/4 since it is written in the form y =mx+b where m is the slope
We want a line that is parallel so it will have the same slope
We know the slope of the new line (3/4) and a point (-1,7)
So we can use the point slope form of the equation
y-y1 = m(x-x1)
y-7 = 3/4(x--1)
y-7=3/4(x+1)
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Which of the following values represent a meaningful difference between two populations? Select all that apply.
Lina20 [59]

Answer: I got you if its not to late  the answer is 2,3

Step-by-step explanation

anything above 2 is meaningful so first you would subtract 21.9 28.1 to get  3.7 its above 2 so that's one do the same for the rest anything above 2 is your answer or right at two really hope this helps

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3 years ago
Find the perimeter of a rectangle length 6yd. 2ft width 4yd 18inches?
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3 0
2 years ago
How do yo work out 9.1 divided by 3575
EastWind [94]
9.1/3575
use long division like this
         _<u>0.</u>_________
3575|91.000000

we find how many of 3575 fit into the 91
it's too big so we go farther
how many 3575 go into 910
too big so we go farther
how many 3575 go into 9100
the answer is 2 so put that in the correct place and mulitply that and put that in the correct place. then we subtract

         _<u>0.</u><u>02</u>_________
3575|91.000000
        -<u>71.50</u>
         19.50
bring down the next number
find how many go into 19.500
the answe ris 5


         _<u>0.</u><u>02</u><u>5</u>_________
3575|91.000000
        -<u>71.50</u>
         19.500
        -<u>17.875</u>
           1.625
          
bring down he next zero ( I fast forward and skip steps for convinience)

                      __
         _<u>0.</u><u>02</u><u>5</u><u>455</u>_________
3575|91.000000
        -<u>71.50</u>
         19.500
        -<u>17.875</u>
           1.6250
          -<u>1.4300</u>
             .19500
            -<u>.17875
</u>                16250
               -14300
and so on untill infinity so the answe ris 0.0254555555555555 (enless fives)
8 0
3 years ago
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
M∠LONm, angle, L, O, N is a straight angle.
Ne4ueva [31]

Answer: 41

Step-by-step explanation:

5 0
3 years ago
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