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3 years ago

2 (y - 5x) - 7y + 9x

Mathematics

2 answers:

pogonyaev3 years ago

7 0

Simplify the expression: -x-5y

liubo4ka [24]3 years ago

4 0

-x-5y I simplified it you should use apps like Socratic and photomath it’s better

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Kim took C dollars to the sporting goods store to spend on new cleats. If the cleats cost $53.62, which of the following express

Mila [183]

C-53.62 times however many she bought.

so c-(53.62x)

because you are taking how much money she spent and subracting it from how much she had

8 0

3 years ago

Please help I promise it is easy algebra 1

Alborosie

Answer:

y=1/x-1/5

Step-by-step explanation:

tell me if u have more questions

y=-x+5

is the line so to make it perpendicular

y=1/x-1/5 i gtg

5 0

3 years ago

-9f +19-1 + f) 12-=8f is this the answer?

jeka94

Hi!

-9f + 19 - 1 + f is actually -8f + 18.

To do this we add like terms, -9f and f; and 19 and -1

-9f + f is -8f

19 - 1 is 18

Therefore your answer is -8f + 18

7 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

3 years ago

The domain of the relation is

vladimir2022 [97]

The domain of a relation is the set of all the x-terms of the relation.

Let's look at an example.

In the image provided I have attached a relation and we want to list the domain.

So, I will list all the x-terms. Notice however that I listed 7 once even though it appears twice in the relation. When listing the domain, you don't repeat the x-terms.

5 0

3 years ago