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3 years ago

AB and BC form a right angle at their point of intersection, B.

Mathematics

1 answer:

const2013 [10]3 years ago

8 0

Answer:

The y-intercept of AB is (4/3) and the equation of BC is y = (-1/6) x + (4/3)

If the y-coordinate of point C is 13, its x-coordinate is 4.

=====================================

Step-by-step explanation:

AB and BC form a right angle at their point of intersection, B.

So, AB ⊥ BC

A = (14 , -1) , B = (2,1)

Part A: Find the y-intercept of AB.

The general equation of the line y = mx + c

Where: m is the slope and c is constant represents y-intercept.

m = (y₂ - y₁)/(x₂ - x₁)

m = $\frac{1 - (-1)}{2-14} = \frac{2}{-12} = - \frac{1}{6}$

y = (-1/6) x + c

Substitute with the point (2,1) to find c

1 = (-1/6) * 2 + c

c = 1 + 2 *(1/6) = 4/3

y = (-1/6) x + 4/3

So, the y-intercept of AB is 4/3

=======================================

Part B: Find the equation of BC.

BC ⊥ AB at point B

if m is the slope of AB, the slope of BC = (-1/m)

m = (-1/6)

So, the slope of BC = (-1/m) = 6

The equation of the line BC ⇒ y = 6x + c

Substitute with the point (2,1) to find c

1 = 6 * 2 + c

c = 1 - 12 = -11

y = 6x - 11

=========================================

Part C: Find x-coordinate of C if y-coordinate of point C is 13.

The equation of the BC is y = 6x - 11

Substitute with the y-coordinate of point C = 13

13 = 6x - 11

6x = 13 + 11 = 24

x = 24/6 = 4

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Use the number line below to find a fraction that is equivalent to 1/2

Yuliya22 [10]

Answer:

4/8

Step-by-step explanation:

There are an infinite number of fractions equal to 1/2, but the number line in the question has 8 spaces, and shows 1 divided into 8 pieces. Therefore, the answer to this question is likely 4/8.

(may I ask why exactly this is put as a college level problem?)

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3 years ago

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faust18 [17]

This question is incomplete because it was not written properly

Complete Question

A teacher gave his class two quizzes. 80% of the class passed the first quiz, but only 60% of the class passed both quizzes. What percent of those who passed the first one passed the second quiz? (2 points)

a) 20%

b) 40%

c) 60%

d) 75%

Answer:

d) 75%

Step-by-step explanation:

We would be solving this question using conditional probability.

Let us represent the percentage of those who passed the first quiz as A = 80%

and

Those who passed the first quiz as B = unknown

Those who passed the first and second quiz as A and B = 60%

The formula for conditional probability is given as

P(B|A) = P(A and B) / P(A)

Where,

P(B|A) = the percent of those who passed the first one passed the second

Hence,

P(B|A) = 60/80

= 0.75

In percent form, 0.75 × 100 = 75%

Therefore, from the calculations above, 75% of those who passed the first quiz to also passed the second quiz.

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3 years ago

On a coordinate plane, 2 lines are shown. Line P Q has points (negative 8, 2) and (4, 2). Line M N has points (8, 6) and (8, neg

DaniilM [7]

Answer:

Slope of PQ = 0

Slope of MN = infinity

PQ and MN are perpendicular to each other

Step-by-step explanation:

for any two points (x1, y1), (x2, y2)given in coordinate plane slope is given by

$y1 - y2/x1-x2\\\\For \ line \ PQ\\slope = 2 - 2/-8-4 = 0\\\\For \ line \ MN \\slope = 6 - (-8)/8-8 = 1/0 = infinity\\\\$

For any line if slope is zero it is parallel to X axis and perpendicular to Y axis

For any line if slope is infinity it is parallel to Y axis and perpendicular to X axis

Also we know X and Y are perpendicular to each other.

Since slope of PQ is zero it is parallel to X axis and perpendicular to Y axis

Since slope of MN is infinity it is parallel to Y axis and perpendicular to X axis.

Thus two lines PQ and MN are perpendicular to each other.

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3 years ago

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Anuta_ua [19.1K]

It’s 112:8 I’m pretty sure:)

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3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

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