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OlgaM077 [116]
3 years ago
5

Using the function p(x)=4x-3 what is p(5)

Mathematics
1 answer:
12345 [234]3 years ago
8 0

Answer:

p(5) = 17

Step-by-step explanation:

p(x) = 4x - 3

p(5) = 4(5) - 3

     = 20 - 3

     = 17

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Three vertices of parallelogram ABCD are A(2,-6),B(-1,2),C(5,3).find the coordinates of vertex D
salantis [7]

Check the first picture below.


for a parallelogram, it has to have two pairs of parallel sides, so BC || AD and AB || CD.

if two lines are parallel, they have the same slope, namely the <u>same rise and run</u>.

notice, the line BC has a run of 6 and a rise of 1, <u>arrows in red</u>, therefore, AD must also have a run of 6 and a rise of 1, so if we move from A 6 units over and 1 up, we'd end up at point D.


now, for part atop

Check the 2nd picture, in a parallelogram, opposite sides are parallel and opposite angles are equal, so if WLP is 144°, then PNW is also 144°.

in a parallelogram, diagonals bisect each other, so each cut the other in halves, so if PM is 9 then MW is also 9 and thus PW is just 9+9=18.


now, about QRST.

notice what we said about a parallelogram above, thus


\bf 8x+13=11x-23\implies 8x+36=11x\implies 36=3x \\\\\\ \cfrac{36}{3}=x\implies \boxed{12=x} \\\\[-0.35em] ~\dotfill\\\\ 2y+12=4y-4\implies 2y+16=4y\implies 16=2y \\\\\\ \cfrac{16}{2}=y\implies \boxed{8=y}


so then TQ ==> 2(8) + 12 ==> 16 + 12 ==> 28.

∡Q ==> 8(12) + 13 ==> 96 + 13 ==> 109°.

and since in a parallelogram, adjacent interior angles add up to 180°, then ∡Q + ∡T = 180°.

∡T ==> 180 - 109 ==> 71°.

3 0
3 years ago
If a point is equidistant from the endpoint of segment, then it is
Mashutka [201]

Hi

.... then it is  located on the perpendicular bisector.

7 0
3 years ago
Is linear or nonlinear?
mr Goodwill [35]

Answer:

Linear

Step-by-step explanation:

7 0
2 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
Al s cat weighs 6 3/5 pounds what is the weight of al,s cat written as a decimal
Tamiku [17]
The correct answer is C
to get 5 to 100, multiply by 20, so multiply 3 by 20 as well. 
8 0
3 years ago
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