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3 years ago

What is the length of XY?

Mathematics

2 answers:

Yuki888 [10]3 years ago

7 0

Answer:

5268

Step-by-step explanation:

length of tangents from a point to the circle are equal.

XY=XZ

91b-283=85b+83

91b-85b=83+283

6b=366

b=366/6=61

XZ=91b-283

=91×61-283

=5551-283

=5268

kherson [118]3 years ago

6 0

Answer:

D. 5267

Step-by-step explanation:

In the triangle XY and XZ are tangent to circle C. This means that the length has of XY and XZ are congruent.

So first you have to set the equations to each other to find what b equals to.

85b+83=91b-283

Add 283 to both sides

85b+366=91b

Subtract 85b on both sides

366=6b

Divide by 6

B=61.

Then you substitute 61 as b in the equation for XY.

85b+83

85(61)+83

5185+83

5268

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PLEASE PLEASE PLEASE HELP ME

lubasha [3.4K]

Answer:

CE = 17

Step-by-step explanation:

∵ m∠D = 90

∵ DK ⊥ CE

∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK

In the two Δ KDE and KCD:

∵ m∠KDE = m∠KCD

∵ m∠DKE = m∠CKD

∵ DK is a common side

∴ Δ KDE is similar to ΔKCD

∴ $\frac{KD}{KC}=\frac{DE}{CD}=\frac{KE}{KD}$

∵ DE : CD = 5 : 3

∴ $\frac{KD}{KC}=\frac{5}{3}$

∴ KD = 5/3 KC

∵ KE = KC + 8

∵ $\frac{KE}{KD}=\frac{5}{3}$

∴ $\frac{KC+8}{\frac{5}{3}KC }=\frac{5}{3}$

∴ $KC + 8 = \frac{25}{9}KC$

∴ $\frac{25}{9}KC - KC=8$

∴ $\frac{16}{9}KC=8$

∴ KC = (8 × 9) ÷ 16 = 4.5

∴ KE = 8 + 4.5 = 12.5

∴ CE = 12.5 + 4.5 = 17

7 0

2 years ago

Manuel hizo un viaje en el coche, en el cual consumió 20 litros de gasolina. El trayecto lo hizo en dos partes: en la primera, c

Alona [7]

Responder:

24 litros; 16 litros; 4 litros

Explicación paso a paso:

Dado que:

Gasolina consumida = 20 litros

Sea la cantidad de gasolina en el tanque = x

Primera parte del viaje = 2/3 de x

Segunda parte del viaje = 1/2 de (x - 2x / 3)

Cantidad de gasolina en el tanque:

2x / 3 + 1/2 (x - 2x / 3) = 20

Solución para x

2x / 3 + x / 2 - x / 3 = 20

(4x + 3x - 2x) / 6 = 20

5 veces / 6 = 20

5 veces = 20 * 6

5 veces = 120

x = 120/5

x = 24

Cantidad de gasolina en el tanque = 24 litros

Litros consumidos en cada etapa:

Primera parte = 2/3 de 24 = 48/3 = 16 litros

2a parte = 0.5 de (24 - 16) = 0.5 * 8 = 4 litros

3 0

2 years ago

Quinn earned $24.50 dog-sitting and $12.70 for recycling cans. Find thetotal amount he earned.

kozerog [31]

Answer: $37.20

Step-by-step explanation:

Just add those 2 amounts together

6 0

3 years ago

Read 2 more answers

X^2/9+2y/7 help me solve

vfiekz [6]

So firstly, we have to find the LCD, or lowest common denominator, of 9 and 7. To do this, list the multiples of 9 and 7 and the lowest multiple they share is going to be your LCD. In this case, the LCD of 9 and 7 is 63. Multiply x^2/9 by 7/7 and 2y/7 by 9/9:

$\frac{x^2}{9}\times \frac{7}{7}=\frac{7x^2}{63}\\\\\frac{2y}{7}\times \frac{9}{9}=\frac{18y}{63}\\\\\frac{7x^2}{63}+\frac{18y}{63}$

Next, add the numerators together, and your answer will be: $\frac{7x^2}{63}+\frac{18y}{63}=\frac{7x^2+18y}{63}$

4 0

2 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

Step-by-step explanation:

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

Hint: sin is only positive in Quadrants I and II

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

Hint: sin and cos are only opposite signs in Quadrants II and IV

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

2 years ago