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Paul [167]
3 years ago
7

I need help. I have to solve for y. 2y+3x=-6

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0

Answer:

-3-3/2x

Step-by-step explanation:

1. Write the problem and subtract -3x on both sides to get the 2y by itself.

2y+3x=-6

    -3x     -3x

= 2y=-6-3x

2. Divide by 2 on each side to get only y completely by itself

= 2y=-6-3x

   /2     /2

= y= -3-3/2x (the -3/2x is supposed to be a fraction.)

hope this helps you understand :)

(if you need more help understanding step by step, type in 'algebra calculator' and click on the first website)

kap26 [50]3 years ago
3 0
The answer is Y=-2
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What is the slope and y-intercept of the line 2x + 3y = 12
sdas [7]

Answer:

2/3 is the slope and 4 is the 4 intersept

Step-by-step explanation:

7 0
3 years ago
In ∆MNP, m∠N = 90º, NH – altitude, m∠P = 21º, PM = 4 cm. Find MH.
Sveta_85 [38]

Answer:

0.51 cm

Step-by-step explanation:

In right triangle MNP, MP = 4 cm, m∠N = 90°, m∠P = 21°

By the sine definition,

\sin \angle P=\dfrac{\text{Opposite leg}}{\text{Hypotenuse}}=\dfrac{MN}{MP}\\ \\MN=MP\sin \angle P\\ \\MN=4\sin 21^{\circ}\approx 1.43\ cm

Now, consider right triangle HMN (it is right because NH is an altitude). By the cosine definition,

\cos \angle M=\dfrac{\text{Adjacent leg}}{\text{Hypotenuse}}=\dfrac{MH}{MN}\\ \\MH=MN\cos \angle M

In the right triangle, two acute angles are always complementary, so

m\angle M=90^{\circ}-m\angle P=90^{\circ}-21^{\circ}=69^{\circ}

Thus,

MH=1.43\cos 69^{\circ}\approx 0.51\ cm

7 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
Linda contructs a triangle with one side 5 inches long and another side 7 inches long.Which is NOT a possible legth for the thir
ladessa [460]
D) 12 would NOT be a possible length for the third side because the sum of the 2 shorter legs should be greater than the longest leg.
8 0
4 years ago
Which expression can be used to find measure angle R
Novay_Z [31]

Answer:

cos-1(-4/14)

Step-by-step explanation:

If

4.5=r

3.5=s

2=t

By law of cosines:

r^{2}=s^{2}+t^{2}  -2st cosR\\\\CosR= \frac{r^{2} -s^{2} -t^{2} }{-2st}  \\\\R=cos^{-1} (\frac{(4.5)^{2}-3.5^{2} -2^{2}  }{-2(3.5)(2)} )\\R= cos^{-1} (-\frac{4}{14}  )

4 0
3 years ago
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