All categories

4 years ago

What must be the length of ZY in order for ZY to be tangent to circle X at point Y?

Mathematics

2 answers:

Paraphin [41]4 years ago

7 0

Not sure but i put 17 units

myrzilka [38]4 years ago

3 0

Answer:

<h2>15 units</h2>

Step-by-step explanation:

In the image attached you can notice that line ZY is tangent at point Y.

Remember that the radius is always perpendicular to tangents, by definition, that means $XY \perp ZY$ .

That means $\triangle XYZ$ is a right triangle where $\angle Y = 90\°$ .

All these facts are deducted form having ZY as a tangent.

We know by given that

$XZ=8+9=17$

$XY=8$ , becaus it's the radius.

Using Pythagorean Theorem

$17^{2} =8^{2}+ZY^{2}$

Solving for $ZY$

$ZY=\sqrt{17^{2} -8^{2} } \\ZY=\sqrt{289-64}\\ ZY=\sqrt{225}\\ ZY=15$

<em>Therefore, the length of ZY must be 15 units to be tangent to circle X.</em>

You might be interested in

A large tank is filled to capacity with 700 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into t

inn [45]

Answer:

The number A(t) of pounds of salt in the tank at time 't' is;

${A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100$

Step-by-step explanation:

In the question, we have;

The volume of pure water initially in the tank = 700 gal

The concentration of brine pumped into the tank = 3 pounds per gallon

The rate at which the brine is pumped into the tank, = 7 gal/min

The rate at which the well mixed solution is pumped out = The same 7 gal/min

The number of pounds of salt in the tank at time 't' is found as follows;

The rate of change in A(t) with time = The rate of salt input - The rate of salt output

$The \ rate \ of \ change \ in \ A(t) \ with \ time = \dfrac{dA}{dt}$

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min

The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min

Therefore, we have;

$\dfrac{dA}{dt} = 21 - \dfrac{A(t)}{100}$

Therefore;

$\dfrac{dA}{dt} + \dfrac{A(t)}{100}= 21$

The integrating factor is $e^{\int\limits {\frac{1}{100} } \, dx } = e^{\frac{x}{100} }$

$e^{\dfrac{t}{100} } \times \dfrac{dA}{dt} + e^{\dfrac{t}{100} } \times\dfrac{A(t)}{100}= e^{\dfrac{t}{100} } \times21$

$\dfrac{d}{dt} \left[ e^{\dfrac{t}{100} } \times{A(t)}{} \right]= e^{\dfrac{t}{100} } \times21$

Using an online tool, we get;

${A(t)}= c_1 \times e^{\dfrac{-t}{100} } + 2,100$

At time t = 0, A(t) = 0

We get;

$0= c_1 \times e^{\dfrac{-0}{100} } + 2,100$

c₁ = -2,100

Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;

${A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100$

8 0

3 years ago

Help me please I need it

Vitek1552 [10]

Answer:

About $4.46

because you can't pay 4.466666667

Step-by-step explanation:

6 people

4 tacos------$5.00

2 burgers--$5.30

4 fries-------$9.00

6 juices-----$7.50

_______________

food ---------$26.80

divide $26.80 by the number of people so 6 then you get 4.466666667 and have to round up

3 0

2 years ago

The amount of time t it takes for a group of volunteers to clean up a park varies inversely with the number of volunteers v. If

Kobotan [32]

Answer:

t = $\frac{10.5}{v}$

Step-by-step explanation:

since t varies inversely with v then the equation relating them is

t = $\frac{k}{v}$ ← k is the constant of variation

to find k use the given condition n = 7, t = 1.5 then

k = vt = 7 × 1.5 = 10.5

t = $\frac{10.5}{v}$ ← is the equation

3 0

3 years ago

Read 2 more answers

7.<br> Solve for x:<br> x^4-5x^2+4 = 0

arlik [135]

Answer:

x=2

Step-by-step explanation:

8 0

3 years ago

Can you please help me please?

VashaNatasha [74]

Answer:

K= -6

Step-by-step explanation:

-12 divided by 2. Hope this helps.

7 0

3 years ago