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Helen [10]
3 years ago
9

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 <u>(4/3)</u> and the equation of BC is y = <u>(-1/6)</u> x + <u>(4/3)</u>

If the y-coordinate of point C is 13, its x-coordinate is <u>4</u>.

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

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)

<u>Part A: Find the y-intercept of AB.</u>

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,  <u>the y-intercept of AB is 4/3</u>

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

<u>Part B: Find the equation of BC.</u>

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

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

<u>Part C: Find x-coordinate of C if y-coordinate of point C is 13.</u>

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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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?)

4 0
3 years ago
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2. (10.03)
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.

3 0
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.

4 0
3 years ago
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PLS HELP WILL MARK BRAINLIEST
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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3 years ago
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