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1 year ago

What is the equation of the line that passes through the point (5, 1) and has a slope of 1/5 ?

Mathematics

1 answer:

OverLord2011 [107]1 year ago

5 0

Answer: $y=\frac{1}{5}x$

Step-by-step explanation:

$y-1=\frac{1}{5}(x-5)\\\\y-1=\frac{1}{5}x-1\\\\y=\frac{1}{5}x$

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Given that x is an integer, what is the solution of |x-1|<img src="https://tex.z-dn.net/?f=%20%5Cleqslant%201" id="TexFormul

Varvara68 [4.7K]

Answer:

see explanation

Step-by-step explanation:

Inequalities of the type | x | ≤ a always have solutions of the form

- a ≤ x ≤ a

Given

| x - 1 | ≤ 1 then

- 1 ≤ x - 1 ≤ 1 ( add 1 to each of the 3 intervals )

0 ≤ x ≤ 2

Since we require an integer solution then x = 0, 1, 2

5 0

2 years ago

Determine the equation of the circle graphed below. ( please help me )

Brums [2.3K]

Answer:

$Equation : \\(x+3)^2 + ( y -2)^2 = 36$

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: Find the centre

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

$Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) = (-3, 2)$

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : Find radius

$Radius = \frac{Diameter }{2}$

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

$Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\$

$= \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12$

Therefore ,

$Radius = \frac{12}{2} = 6$

Step 3 : Equation of the circle

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

$(x - h)^2+(y -k)^2 = r^2$

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

$(x - (-3))^2 + (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36$

5 0

2 years ago

What is the relationship between the discriminant of a quadratic and its graph?

faltersainse [42]

Answer:

In a quadratic equation of the shape:

y = a*x^2 + b*x + c

we hate that the discriminant is equal to:

D = b^2 - 4*a*c

This thing appears in the Bhaskara's formula for the roots of the quadratic equation:

$x = \frac{-b +-\sqrt{b^2 - 4ac} }{2a}$

You can see that the determinant is inside a square root, this means that if D is smaller than zero we will have imaginary roots (the graph never touches the x-axis)

If D = 0, the square root term dissapear, and this implies that both roots of the equation are the same, this means that the graph touches the x axis in only one point, wich coincides with the minimum/maximum of the graph)

If D > 0 we have two different roots, so the graph touches the x-axis in two different points.

4 0

2 years ago

Find the gradient and y intercept of 3x+4y=5

noname [10]

Answer:

gradient = - $\frac{3}{4}$ , y- intercept = $\frac{5}{4}$