What is the coefficient of 18 ?t=24

1 answer:

8 0

I assume by the question mark that you are asking is 18 is the coefficient, If that is what you mean than you are right the coefficient is the number thetas with the variable for example 6x+y=14 the coefficient is 6 because it accompanies a variable.

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Megan went to the store and bought 4 notebooks and 9 pencils for a total of $8.25. Joanne went to the same store and bought 7 no

Rzqust [24]

Let notebooks be x and let pencils be y

4x+9y=8.25 (equation 1)
7x+14y=14.00 (equation 2)

Multiply equation 1 by 7
So that you get 28x+63y=57.75

Multiply equation 2 by 4
So that you get 28x+56y=56.00

28x+63y=57.75 -
28x+56y=56.00
-----------------------
7y=1.75

Divide both sides by 7
1.75/7 is 0.25

So, now we know that one pencil costs 0.25 cents

Sub this y value into equation one

4x+9(0.25)=8.25

4x+2.25=8.25

Move 2.25 to the other side as a negative number

4x= 6

Divide both sides by 4

6/4 is $1.50

So, one pencil is equal to 0.25 cents and one notebook is equal to $1.50

5 0

3 years ago

PLEASE HELP! URGENT WILL GIVE BRAINLEST! PLEASE ANSWER ALL QUESTIONS!!! IF YOU DON'T I WILL REPORT.

Nesterboy [21]

Answer:

1. b) 135°, 225°

2. c) 0°, 180°

3. a) π/2

4. b) 30°, 150°, 210°, 330°

5. a) 1.25 radians and 4.39 radians

Step-by-step explanation:

Question 1

$\begin{aligned}\cos \theta & = -\dfrac{\sqrt{2}}{2}\\\implies \theta & = \cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)\\\\& = 135^{\circ} \pm 360^{\circ}n, 225^{\circ} \pm 360^{\circ}\\\\& = 135^{\circ}, 225^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}$

Question 2

$\begin{aligned}\cos^2 \theta-1 & = 0\\\implies \cos^2 \theta & = 1\\\cos \theta & = \pm1\\\theta & = \cos^{-1}(\pm1)\\& = 0^{\circ} \pm 360^{\circ}n, 180^{\circ} \pm360^{\circ}n\\& = 0^{\circ}, 180^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}$

Question 3

$\begin{aligned}\sin \theta & = 1\\\implies \theta & = \sin^{-1}(1)\\\theta & = \dfrac{\pi}{2}\pm2\pi n\\\theta & = \dfrac{\pi}{2}\quad \textsf{for}\:0 \leq \theta < 2 \pi \end{aligned}$

Question 4

$\begin{aligned}3 \tan^2 \theta-1 & = 0\\\implies \tan^2 \theta & = \dfrac{1}{3}\\\tan \theta & = \pm\dfrac{1}{\sqrt{3}}\\\theta & = \tan^{-1}\left(\pm\dfrac{1}{\sqrt{3}}\right)\\& = 30^{\circ}\pm180^{\circ}n, 150^{\circ}\pm180^{\circ}n\\& = 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}$

Question 5

$\begin{aligned}2 \tan \theta -6 & = 0\\\implies \tan \theta & = 3\\\theta & = \tan^{-1}(3)\\\theta & = 1.25\pm \pi n\\\theta & = 1.25, 4.39 \quad \textsf{for}\:0^{\circ}\leq \theta < 2 \pi \end{aligned}$