1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ksivusya [100]
3 years ago
13

Solve -15-2g+6g=1+6g

Mathematics
2 answers:
katrin [286]3 years ago
7 0
<span>-15-2g+6g=1+6g
Combine like terms
-2g=16
Divide by -2 on both sides
g=-8</span>
AleksAgata [21]3 years ago
7 0
<span>-15-2g+6g=1+6g

First: combine like terms
You'll get: -15+4g=1+6g
Next: You would subtract 4g on both sides to where you can still get all of your like terms together
You'll get: -15=1+2g
Then: You'll subtract 1 on both sides because you are still getting your like terms together
You'll get: -16=2g
Finally: You'll divide each side by 2, so your g would be alone
You'll get: -8=g <That would be your answer

HOPE THIS HELPS! ^_^</span>
You might be interested in
The table below shows some prices at a produce stand.
Rainbow [258]
Well she has $10 and each spinach is $3 so she can not but no more than 3.
3 0
3 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

7 0
1 year ago
Please help me understand, ill give brainliest!
Mkey [24]

Answer:

C 58

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
The graph shows the amount of money paid when purchasing bags of candy at the zoo:
s2008m [1.1K]

Answer:

$2 because the x axis is bags of candy 1 and the y axis is the cost 2 therefore 1 bag of candy cost $2 and 2 bags of candy cost $4

6 0
3 years ago
Coach Evans recorded the height, in inches of each player on his team. The results are shown.
marysya [2.9K]

Answer:

3

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile (Q_1)

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile (Q_3)

Terms in arranged in ascending order:

57,60,60,61,62,62,63,63,64

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

Q_1 is median of terms 57,60,60,61

Number of terms (n) = 4

Median = \frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th}  }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60

So, Q_1=60

So, 25% of the heights of a team is less than 60 inches

Q_3 is the median of terms 62,63,63,64

Median = \frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th}  }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63

So, Q_3=63

So, 75% of the heights of a team is less than 63 inches

Interquartile range = Q_3-Q_1=63-60=3

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

4 0
3 years ago
Other questions:
  • You are standing 45 meters from the base of the Empire State Building. You estimate that the angle of elevation to the top of th
    6·1 answer
  • Which statement is true?
    8·1 answer
  • An artist receives 20% royalty on the retail price of his recordings. If he receives $36,000 in royalties, what is the dollar am
    8·1 answer
  • Find a solution to the following system of equations.<br> 5x + y = 10 <br> 4x + 5y = 8
    7·1 answer
  • Which algebraic expressions are polynomials
    10·1 answer
  • I NEED HELP ASAP it’s due today!
    13·1 answer
  • What does -4(3x-4)=44<br> What is x
    7·1 answer
  • PLEASE HELP I WILL GIVE BRAINLIEST TO THE FIRST PERSON THAT ANSWER PLEASE
    11·1 answer
  • The area of a square is 275 square meters and one side length is 55 meters. What is the perimeter of the square?
    14·1 answer
  • the rectangular post shown at the right measures 2w-30 by w. the poster has an area of 5400 cm^2. what is The value of w ?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!