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3 years ago

RS is the perpendicular bisector of TU. Find x and y

Mathematics

1 answer:

german3 years ago

8 0

Answer:

x = 1

y = 30°

Step-by-step explanation:

is RS is ⊥ to TU then TS = US so we can use this equation to find 'x':

20x-3 = 5x+12

15x = 15

x = 1

we also know that RS intersecting with TU creates a 90° angle

if 3y = 90 then y = 30

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A line has an undefined slope and includes the points (n, - 5) and (4,1). What is the value of n?

Shalnov [3]

Answer:

n=4

Step-by-step explanation:

To get an undefined slope, the denominator has to be zero

m= (y2-y1)/ (x2-x1)

This means that x2-x1 = 0

or x2=x1

So 4=n

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I need to know how to write a proof for why CF is the perpendicular bisector of BD

Soloha48 [4]

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3 years ago

1. Suppose a and b are real numbers where a > 0 and b < 0.

Nat2105 [25]

a) –a + b is negative

b) a – b is positive

c) b-a is negative

Step-by-step explanation:

1. Suppose a and b are real numbers where a > 0 and b < 0.

a. Is –a + b positive or negative. Explain how you know.

We know a > 0 and b < 0. so, let a =6 and b = -5

Putting values:

–a + b

= -(6)+(-5) = -6-5 = -11

So, –a + b is negative

b. Is a – b positive or negative? Explain how you know.

We know a > 0 and b < 0. so, let a =6 and b = -5

Putting values:

a - b

=6-(-5) = 6+5 = 11

So, a – b is positive

c. Is b – a positive or negative. Explain how you know

We know a > 0 and b < 0. so, let a =6 and b = -5

Putting values:

b-a

=(-5)-(6)

= -5-6

= -11

So, b-a is negative

Keywords: Solving Integers:

Learn more about solving integers at:

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3 years ago

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$