Enable contrast version

Tutor profile: Gabby N.

Inactive
Gabby N.
Masters Degree in Education with a Bachelor of Arts in Mathematics
Tutor Satisfaction Guarantee

Questions

Subject: Geometry

TutorMe
Question:

What type of single transformation was applied to quadrilateral A to get quadrilateral B, if B is exactly the same as A except for being smaller where the length between points has decreased? Was the transformation a translation, rotation, reflection or dilation

Inactive
Gabby N.
Answer:

The quadrilateral is not describe as being shifted around the space, thus it is not a translation. It was not described as having been moved around a single point, thus it was not rotated. It was not described as being reflected over a certain line, thus it is not a reflection. All three of these types of transformations are rigid transformations which means the length between points are preserved. A non-rigid transformation is does not need the lengths between points to be preserved thus we can scale up or scale down a shape. This scaling up to increase and scaling down to decrease in size is known as dilation. Quadrilateral A was transformed by dilation to become quadrilateral B.

Subject: Pre-Algebra

TutorMe
Question:

On a straight line there are two rays that start at the same point making three different angles. One angle is a right angle, one is 30 degrees and the other is equal to 2x degrees. What is the value of x ?

Inactive
Gabby N.
Answer:

If the angles are all connecting on a straight line, we know that their sum is 180 degrees. We can then write an equation based on our given information. We know that one angle is a right angle which is 90 degrees, one is 30 degrees and one has a variable: 2x degrees. Our equation is then the sum of these three angles set equal to 180: 90 + 30 + 2x = 180. We now need to add like terms 120 + 2x = 180 Subtract 120 from both sides of the equation to get 2x by itself 2x= 60 We now know that our missing angle is 60 degrees but the question asked what the value of x was so we need to now divide both sides of the equation to know that x=30 We can check our work by taking our original equation and plugging in 30 for x 90 +30 + 2(30)= 180 90 + 30 +60 = 180 YES!

Subject: Algebra

TutorMe
Question:

Using Systems of equations with elimination or substitution, solve the following word problem. In the barn there were 35 heads of horses and chickens. If there were 110 legs in the barn, how many horses and chickens where in the barn? Please show your work with your answer.

Inactive
Gabby N.
Answer:

To begin we can write our system of equations with the given information. Because there are two unknown variables we will need two equations to solve this problem. Let H represent horses and C represent chickens. We can then write our first equation as the number of horses and the number of chickens adding to equal the total number of heads found in the barn; thus, our first equation should look like H + C = 35. Our second equation is a little more complicated and for this problem we will assume that horses have 4 legs and chickens have 2 legs. Based on the information given in the problem we know that the number of horse legs and the number of chicken legs adds to equal 110 legs total. When writing our second equation, we will multiply 4 times H because we know that for every horse in the barn there are 4 legs attached. We can then add 4 times H to 2 times C because we also know that for every chicken there are 2 legs. If we had 4 times H and 2 times C, we will get the total number of legs; thus our second equation is 4H + 2C = 110. Now that we have our system of equations: H + C = 35 4H + 2C = 110 We can choose to solve this with either elimination or substitution. Solving with Substitution: We must choose if we want to find the number horses or chickens first. For this example, we will solve for horses first. Taking the first equation H + C=35, we will solve for C in terms of H. This means we want C on one side of the equation by itself. We can do this easily by subtracting H from both sides of the equation. Giving us C = 35 - H as our new equation! We will now take the second equation 4H +2C =110 and substitute C with the new equation in terms of H. It should now look like 4H + 2(35-H)=110. Our next step is to simplify and add like terms in this equation: 4H + 2(35-H)=110 4H +70 -2H = 110 step 1: multiply 2 times what is in the parentheses 2H +70=110 step 2: add all H variables 2H= 40 step 3: subtract 70 from both sides of the equation to combine like terms H=20 step 4: divide both sides of the equation by 2 to get H by itself. Now that we know there are 20 horses, we can then plug the value 20 into one of the original equations or into the one we manipulated (C =35-H ) to find out how many chickens there are. C=35 - H if H= 20 then C=35-20 thus C= 15 there are 15 chickens. To Check our work, simply take our answers, H=20 and C=15 and plug them back into our two equations. If we found the correct answers the equations should be true: H + C = 35 Check: 20 +15=35 YES! 4H + 2C = 110 Check: 4(20) +2 (15) = 110 simplified 80+30=110 YES! If you choose to solve this problem with Elimination, below is your solution: H + C = 35 4H + 2C = 110 choosing to solve for C first this time, we need to eliminate H from the system of equations. Step one: multiply the whole first equation by -4 to get -4H -4C=-140 the system of equation now looks like -4H -4C= -140 4H + 2C = 110 Step two: Add both equations together combining only like terms on each side of the equation to get: 0H -2C = -30 thus H is eliminated leaving us with -2C = - 30 step three: divide both sides of the equation by -2 to find that C= 15. Step four: go back to one of the original equations plug in the value of C and solve for H H + C = 35 H + 15 = 35 H= 20 You can then check your work the same way we did when solving it the first time.

Contact tutor

Send a message explaining your
needs and Gabby will reply soon.
Contact Gabby

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2021 TutorMe, LLC
High Contrast Mode
On
Off