All categories

3 years ago

What is the equation of a line that passes

Mathematics

1 answer:

OleMash [197]3 years ago

4 0

Answer:y=9.5x-0.5

Step-by-step explanation:

Ok, so you need to use the equation y2-y1/x2-x1 to find the slope. So (18.5-4.25)/(2-0.5) so 14.25/1.5 , calculate that and you get 9.5 which is your slope, apply that to the slope formula which is y=mx+b where (m) is slope and (b) is y intercept and you get y=9.5x-0.5

You might be interested in

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

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

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

3 years ago

Write the equation in slope-intercept form then change it to standard form with integer coefficients.

SCORPION-xisa [38]

Answer:

1. Slope-intercept form: $y = 2x - 4$

Standard form: $2x - y = 4$

2. Slope-intercept form: $y = \dfrac12x - 1$

Standard form: $x - 2y = 2$

3. Slope-intercept form: $y = \dfrac45x + \dfrac{21}{5}$

Standard form: $4x -5y = - 21$

Step-by-step explanation:

Slope intercept form: $y = mx + b$

where:

$y$ = y-coordinate
$m$ = slope
$x$ = x-coordinate
$b$ = y-intercept

Standard form: $Ax + By = C$

$\textsf{1. as} \ m=2: \ \ y = 2x + b$

$\textsf{at} \ (6, 8): \ \ 8 = 2(6) + b$

$\implies b = -4$

Slope-intercept form: $y = 2x - 4$

Standard form: $2x - y = 4$

$\textsf{2. as} \ \ m = \dfrac12: \ \ y = \dfrac12x + b$

$\textsf{at} \ (4, 1): \ \ 1 = \dfrac12(4) + b$

$\implies b = -1$

Slope-intercept form: $y = \dfrac12x - 1$

Standard form: $x - 2y = 2$

$\textsf{3. as} \ \ m = \dfrac45: \ \ y = \dfrac45x + b$

$\textsf{at} \ (1,5): \ \ 5 = \dfrac45(1) + b$

$\implies b = \dfrac{21}{5}$

Slope-intercept form: $y = \dfrac45x + \dfrac{21}{5}$

Standard form: $4x -5y = - 21$

7 0

2 years ago

Limit as x approaches infinity: 2x/(3x²+5)

Nonamiya [84]

$\bf \lim\limits_{x\to \infty}~\cfrac{2x}{3x^2+5}\implies \cfrac{\lim\limits_{x\to \infty}~2x}{\lim\limits_{x\to \infty}~3x^2+5}$

now, by traditional method, as "x" progresses towards the positive infinitity, it becomes 100, 10000, 10000000, 1000000000 and so on, and notice, the limit of the numerator becomes large.

BUT, notice the denominator, for the same values of "x", the denominator becomes larg"er" than the numerator on every iteration, ever becoming larger and larger, and yielding a fraction whose denominator is larger than the numerator.

as the denominator increases faster, since as the lingo goes, "reaches the limit faster than the numerator", the fraction becomes ever smaller an smaller ever going towards 0.

now, we could just use L'Hopital rule to check on that.

$\bf \lim\limits_{x\to \infty}~\cfrac{2x}{3x^2+5}\stackrel{LH}{\implies }\lim\limits_{x\to \infty}~\cfrac{2}{6x}$

notice those derivatives atop and bottom, the top is static, whilst the bottom is racing away to infinity, ever going towards 0.

5 0

3 years ago

What is the area of the shaded sector

Whitepunk [10]

Answer:

Area = 634.21 square metre

Step-by-step explanation:

Area of shaded region = Area of the circle - Area of the sector

$Area \ of \ the \ circle = \pi r^2 = 225 \pi\\\\Area \ of \ the \ sector = \pi r^2 \frac{ \theta}{360} = \pi \times 225 \times \frac{37}{360}$