TutorMe homepage
Subjects
PRICING
COURSES
Start Free Trial
Eslam A.
Mathematics and Physics Student - Tutor for 4 years
Tutor Satisfaction Guarantee
Trigonometry
TutorMe
Question:

The area of a right triangle is 50. One of its angles is 45 degrees. Find the lengths of the sides and hypotenuse of the triangle. (i.e. solve this triangle)

Eslam A.

In this problem we need to solve the triangle to get the measurements of all angles and lengths of all sides, we will analyze what we have to proceed correctly through the question. We have a right-angled triangle. so the biggest angle here is 90 degrees. we know that the sum of angles of any triangle is 180 degrees, and it's given that we have a 45 degrees angle, so the last angle will be 180 - the sum of the two other angles so the measurement of the third angle is: 180 - (90 + 45) = 180 - 135 = 45 degrees, (take care of the order of arithmetic operations, first compute the bracket then the subtraction) we note here that the two angles have the same measurement, so the triangle is isosceles, and the length of the two sides other than the hypotenuses are equal. no we will use the rule of area of the triangle, but we will adjust it to have one simple equation to solve, so we will use the right angle and the two sides to plug into the relation, as the relation says: area = $$1/2$$ the product of the two sides * since the included angle let the side length be L, so area = $$1/2*L*L*\sin(90)$$ = $$1/2 L^{2} \sin(90)$$, solve this while the area is 50 square unit and $$\sin(90) = 1$$ so $$50 = 1/2 L^{2} *1$$ , rearrange : $$100 = L^{2}$$ , so $$L = \sqrt(100)$$ = 10 units (we excluded the negative result because there is no length in negative) no we have the length of two sides and we miss the hypotenuse, we either use Pythagoras theorem to get it or use the sine or cosine of any 45-degree angle we will use the sine of the 45 degrees angle, we know that sine = $$opposite/hypotenuses$$, so $$\sin(45)=L/h$$, rearrange to get h, so $$h=L/\sin(45)$$=$$10/\sin(45)$$=$$10 ÷ 1/\sqrt(2)$$ = $$10*\sqrt(2)$$, so the hypotenuse = $$10\sqrt(2)$$ we need to check using Pythagoras theorem, $$h^{2} = L^{2} + L^{2} = 2*L^{2} = 2*100 = 200$$ while $$h^{2}=(10\sqrt(2))^{2}=200$$ also, so we're correct. The hypotenuse = $$10\sqrt(2)$$ and the other sides are equal and their length is 10 units

Calculus
TutorMe
Question:

There are 50 apple trees on a farm, each tree produces 800 apples. For each additional tree planted in the farm, the output per tree drops by 10 apples. How many trees should be added to the existing farm in order to maximize the total output of trees?

Eslam A.

The problem here is a maxima and minima one, we need to find the maximum number of trees we can add to the existing farm while taking into consideration the drop down in the number of apples per tree to get the maximum income. First, we shall get a formula for the total output of the farm, then we shall differentiate this product and equate with zero to get the possible answers. Let the product of the farm be P, and the extra number of trees is x, so in case there were no extra trees, the product is $$50*800$$, but for each tree more than 50 (which will be $$50+x$$) the product per tree will reduce by 10 per each tree (so the product will be $$800-10 x$$), so the net product P will be $$(50+x)*(800-10 x)$$ and this is our formula that we will work with. So P = $$(50+x)*(800-10 x)$$ expand : $$P = 50*800+50*(-10 x)+x*800-x*(-10 x)$$ then rearrange : $$P = -10 x^2 + 300 x + 40000$$ we need to maximize it, so we differentiate with respect to x, then equate with zero differentiate $$dP/dx = -20 x + 300$$ set $$dP/dx = 0$$ $$20 x = 300$$ so $$x = 300/20 = 15$$ we need to check if this number is the maximum value or not, so we take the second derivative of P with respect to x, if it's positive, so this number will lead to minimum, if negative, this number will lead to maximum. $$d^{2}P/dx^{2} = -20$$ (negative) so it's the maximum of P we want to make another check, so we will check some number of x less than 15 and more than 15. for $$x = 15 : P = 42250$$ apple for $$x = 20 : P = 42000$$ apple for $$x = 10 : P = 42000$$ apple So we are correct.

Algebra
TutorMe
Question:

Find the equation of the line that passes through the points $$(-1,-1)$$ and $$(1,1)$$

Eslam A.

In this question, we need to find a generic equation for a straight line that passes through these two points using only these two points. We have many ways to find the equation of a straight line, but when we have two points, it's more convenient to use this formula : $$(y - y_{1})/(x - x_{1}) = (y_{2} - y_{1})/(x_{2} - x_{1})$$, where $$y$$ and $$x$$ are the coordinates of any point, and $$(x_{1}, y_{1})$$ are $$(-1,-1)$$, and $$(x_{2}, y_{2})$$ are $$(1,1)$$, and the expression $$(y_{2} - y_{1})/(x_{2} - x_{1})$$ is the slope of this line now plug in the numbers, taking into consideration the signs of each number $$(y - (-1))/(x - (-1)) = (1 - (-1))/(1 - (-1))$$, simplifying this equation : $$(y + 1)/(x + 1) = (1 + 1)/(1 + 1)$$ = $$(y + 1)/(x + 1) = (2)/(2) = 1$$ so $$(y + 1)/(x + 1) = 1$$, rearrange again, $$y+1=x+1$$, we need to simplify, so we cancel $$+1$$ from each side, leaving us with $$y=x$$ which is our equation, and this is somehow intuitive, as the two points are opposite to each other with the same value, so the equation of the straight line should pass by the origin and has no interception with $$x$$ or $$y$$ we need to check this answer, so we will check if the two points satisfy this equation or not for the first point $$-1 = -1$$ which is correct for the second point, $$1=1$$ which is also correct. so the equation $$y=x$$ is correct.

Send a message explaining your
needs and Eslam will reply soon.
Contact Eslam
Ready now? Request a lesson.
Start Session
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 Session" 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.