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2 years ago

Find the slope and y intercept

Mathematics

1 answer:

alexgriva [62]2 years ago

7 0

Slope: 5/1 (5 if simplified)
y-intercept: 10

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Fine the equation, in standard form, of the line passing through the points (3,-4) and (5,1)

Talja [164]

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (3, -4) and (5, 1). Substitute:

$m=\dfrac{1-(-4)}{5-3}=\dfrac{5}{2}\\\\y-(-4)=\dfrac{5}{2}(x-3)$

The standard form: $Ax+By=C$

$y+4=\dfrac{5}{2}(x-3)$ multiply both sides by 2

$2y+8=5(x+3)$ use distributive property

$2y+8=5x+15$ subtract 2y from both sides

$8=5x-2y+15$ subtract 15 from both sides

$-7=5x-2y$

<h3>Answer: 5x - 2y = -7</h3>

5 0

2 years ago

4x + 3 = x + 2(x + 7) x=

stiv31 [10]

Answer:

X= -11/4 + -1/4 √145 or X= -11/4 + 1/4 √145

Step-by-step explanation:

Hope this helped<3

4 0

2 years ago

−4(m+3)=24 What does m=

Ket [755]

Answer: m= -9

Step-by-step explanation: First use distributive property, -4(m) is -4m and -4(3) is -12. Then you have -4m -12 =24. Then use 12 and add it to the other side of the equation, cause of the subtraction sign, 12+24=36. Then you have -4m=36. Then divide the 4 to get the m by its self. 36 divided by -4 is -9. Your answer is m=-9

Hope this helps :)

5 0

2 years ago

I need help please !!!

Zarrin [17]

Answer:

1. it is given

2. the definition of angle bisector

3. the reflexive property

4. SAS

5. corresponding parts of congruent triangles are congruent

3 0

2 years ago

Really need help guys! Love you if you can solve them wiz steps!!!!

Marta_Voda [28]

Answer:

Step-by-step explanation:

$x^2+y^2-8x+4y+4=0\\\\a)\\\\x^2-8x+y^2+4y+4=0\\\\x^2-2*x*4+(y^2+2*y*2+2^2)=0\\\\x^2-2*x*4+4^2+(y+2)^2=4^2\\\\(x-4)^2+(y+2)^2=4^2\\$

Hence,

The radius of the circle is 4 units, coordinates of its centre are (4,-2).

$b)\\\\y=0\\x^2+0^2-8x+4*0+4=0\\x^2-8x+4=0\\a=1\ \ \ \ b=-8\ \ \ \ c=4\\D=(-8)^2-4*1*4\\D=64-14\\D=48\\\sqrt{D}=\sqrt{48} \\ \sqrt{D}=\sqrt{16*3} \\\sqrt{D}=\sqrt{4^2*3} \\\sqrt{D}=4\sqrt{3} \\\displaystyle\\x=\frac{-(-8)б4\sqrt{3} }{2*1} \\\\x=\frac{8б4\sqrt{3} }{2} \\x_1=4-2\sqrt{3} \\x_2=4+2\sqrt{3}$

$c)\\\\A(6,2\sqrt{3}-2)\\ (x-4)^2+(y+2)^2=4^2\\(6-4)^2+(2\sqrt{3} -2+2)^2=16\\2^2+(2\sqrt{3} )^2=16\\4+2^2*(\sqrt{3})^2=16\\ 4+4*3=16\\4+12=16\\16\equiv16$

$d)\\A(6,2\sqrt{3}-2)}\\\sqrt{3} x+3y=\\\sqrt{3}(6)+3(2\sqrt{3} -2)=\\ 6\sqrt{3}+6\sqrt{3} -6=\\ 12\sqrt{3}-6$

5 0

11 months ago