Ask question

Ask question

All categories

3 years ago

11

4. What is the equation of a line with a slope of 5 that passes through the point (4,2) ? Express this equation in all three for

ms: point-slope, slope-intercept, and standard.

2 answers:

Colt1911 [192]3 years ago

5 0

Incase you did not know, f(x) is a way of writing equations, for lines, if you do not know what the equation is.

2 = f(4)
f(x) = 5x-c
f(4) = 20-c
c = 20-2 = 18
f(x) = 5x-18

thus...
y = 5x-18

also...
the y intercept is -18

stiv31 [10]3 years ago

3 0

Slope = 5
y = mx + b
2 =5(4) + b
b = 2 - 20
b = -18

<span>point-slope equation
y - 2 = 5(x -4)

</span><span>slope-intercept equation
y = 5x - 18

standard form
5x - y = 18</span>

You might be interested in

Which of the binomials below is a factor of this trinomial? 6x2 - 6x - 72

vova2212 [387]

Just factor to find the roots

<span>6x² - 6x - 72 = 0 </span>
<span>6(x² - x - 12) = 0 </span>
<span>6(x - 4)(x + 3) = 0 </span>

<span>B: x - 4</span>

3 0

3 years ago

Read 2 more answers

Algebraic equation of 185 is less than a number

kkurt [141]

Answer:

n-185

Step-by-step explanation:

since 185 needs to be subtracted from a number, or n

4 0

3 years ago

Write the fractions in order from least to greatest -3 3/4, -3 17/25, -3 13/20, -33/5

Natali [406]

3 17/25 3 13/20 3 3/5 3 3/4

7 0

3 years ago

Michael writes down 4 different factors of 60

Marizza181 [45]

Answer:

34+(-12) =22

Step-by-step explanation:

because they said that adding numbers but they didn't said that negative Or positive. so i took those numbers

3 0

3 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

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

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

<em>Hence, circle Q has a smaller radius</em>

4 0

3 years ago