The equation of a circle is (X-3)² + (Y+5)²=25

4 /5 fraction can be rewritten as _____

patriot [66]

Answer:

8/10, 12/15, 16/20 and so on

Step-by-step explanation:

4x2/5x2 = 8/10
4x3/5x3 = 12/15
4x4/5x4 = 16/20 and so on

3 0

2 years ago

. Which algebraic expression is equivalent to the expression below?

I am Lyosha [343]

Since you are not distributing at all, you can ignore the parenthesis :)

When ignoring the parenthesis your equation is
18x + 11 - 4x

Then just combine like terms :)

14x + 11

Your answer is C.

Hope this helped!

5 0

3 years ago

Please help this is due in tommorow! I will mark you as brainliest!

pshichka [43]

Answer:

325.26

Step-by-step explanation:

that's the answer because you have to multiply the top

I have question do we have to do something to the bottom?

5 0

3 years ago

Lucy saved $50 by buying a chair that was on sale for 25% off the regular

neonofarm [45]

Answer:

$200

Step-by-step explanation:

Represent the regular price of the chair by c.

Then 0.25c =$50, and c = $200.

The regular price, c, was $200.

4 0

3 years ago

Read 2 more answers

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

3 years ago