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3 years ago

How many solutions are there to the following system of equations?

Mathematics

1 answer:

blagie [28]3 years ago

5 0

Lets simplify these equations into slop-intercept form (y=mx+b)

When we do that, the equations turn to

y=-6x/24+10/6 or y=-1/4*x+5/3
y=-2x/8+7/8 or y=-1/4*x+7/8

When you do this, you see that the slope (m) is the same. This means that they are parallel lines. That means these lines will never meet, which means there will be no solution

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Plzzzzzzzz helpe out

Dvinal [7]

Answer:

$\frac{cos-sin^{2}+1 }{sin(1+cos)}\\\\=\frac{cos - sin^{2}+sin^{2}+cos^{2} }{sin(1+cos)}\\\\=\frac{cos+cos^{2} }{sin(1+cos)}\\\\=\frac{cos(1+cos)}{sin(1+cos)}\\\\=\frac{cos}{sin}\\\\=cot$

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Step-by-step explanation:

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2 years ago

Read 2 more answers

(9x2 - 8x + 4) - (4x2 + 3x - 9) A. 13x2 - 11x + 5 B. 13x2 - 11x + 13 C. 5x2 - 5x + 13 D. 5x2 - 11x + 13

Troyanec [42]

The answer is d) 5x2-11x+13

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2 years ago

Which pair of numbers has an LCM of 8

maxonik [38]

Answer:

2,8

Step-by-step explanation:

3 0

2 years ago

Angl is a square. The equation of line NG is Y= 1/2 X - 6. Find the equation of line LG in slope intercept form given that the c

Irina18 [472]

Answer: y=-2x-9

Step-by-step explanation:

If ANGL is a square, then NG and LG are adjacent sides.

Adjacent sides are perpendicular. [Each angle is 90°]

The equation of line NG is $Y=\dfrac12 X-6$ .

By comparing it to equation in slope intercept form y=mx+c ( where , m= slope , c=y-interecpt)

slope = $\dfrac12$

Let slope of LG be <em>n</em>, then

$n\times \dfrac{1}{2}=-1$ [Product of slopes of two perpendicular line =-1]

$\Rightarrow n=-2$

Equation of a line passes through (a,b) and have slope m is given by :-

$(y-b)=m(x-a)$

Equation of LG :

$(y-1)=-2(x-(-5))\\\\\Rightarrow\ y-1=-2x-10\\\\\Rightarrow\ y=-2x-9$ [In intercept form]

3 0

2 years ago

What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE

Nutka1998 [239]

Answer:

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),$ $\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

Step-by-step explanation:

Let points D, E and F have coordinates $(x_D,y_D),\ (x_E,y_E)$ and $(x_F,y_F).$

1. Midpoint M of segment DF has coordinates

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).$

2. Midpoint N of segment EF has coordinates

$\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0

2 years ago