Question about Super Tutor Trigonometry (ESDTRIG) for PC

C = 3B = 2(A+B)

3/2 B = A+B

1/2B = A

C = 2 (1/2B + B)

C = 3B

**A = 1/2B

C = 3B

B = B

1/2B + 3B + B = 4 1/2 B

180/4.5B => B = 40

If B = 40, then A = 1/2B = 20 and C = 3B = 120.

**answer: A = 20 degrees, B = 40 degrees and C = 120 degrees.

Posted on Jan 10, 2009

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Posted on Jan 02, 2017

30 and 80.

If this is homework, be sure to show your work.

If this is homework, be sure to show your work.

Nov 15, 2016 | Office Equipment & Supplies

I find the easiest way to solve these is to sketch them first (I'm a visual learner;) We get a nice right-angled triangle, with the right-angle at B. The formula for the area of a triangle is 1/2 * base* height or (base * height)/2.

We can use BC or AB as the base.

If we use BC as the base, the length is 9-4 or 5. The height is 6-2 or 4.

We can now but the base and the height in the formula to figure out the area.

Good luck.

Paul

We can use BC or AB as the base.

If we use BC as the base, the length is 9-4 or 5. The height is 6-2 or 4.

We can now but the base and the height in the formula to figure out the area.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

The area of a triangle is 1/2 times base times height. A sketch of the triangle in the coordinate plane will determine how easy or hard this will be to be. From the sketch, you will see that this is a right-angled triangle with B being the right-angle. This makes it easier because we can easily determine the base and the height to use in the formula.

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

The area of a triangle is 1/2 times base times height. A sketch of the triangle in the coordinate plane will determine how easy or hard this will be to be. From the sketch, you will see that this is a right-angled triangle with B being the right-angle. This makes it easier because we can easily determine the base and the height to use in the formula.

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

To solve a system of equations, there are two algebraic ways to solve them, one is by substitution, while the other is elimination.

1) 2a + 3b + 5 = 0

2) 2a + 4b + 10 = 0

-----------------------

-b -5 = 0 (subtract 2 from 1)

+b-b-5 = b

-5 = b

Substitute into 1) to solve for a.

2a + 3(-5) + 5 = 0

2a -15 + 5 = 0

2a -10 = 0

2a -10 + 10 = 10

2a = 10

---- ----

2 2

a=5

Substitute into 2) to check if we did it correctly.

2(5) + 4(-5) + 10 = 0

10 -20 + 10 = 0

0=0

Good luck.

Paul

1) 2a + 3b + 5 = 0

2) 2a + 4b + 10 = 0

-----------------------

-b -5 = 0 (subtract 2 from 1)

+b-b-5 = b

-5 = b

Substitute into 1) to solve for a.

2a + 3(-5) + 5 = 0

2a -15 + 5 = 0

2a -10 = 0

2a -10 + 10 = 10

2a = 10

---- ----

2 2

a=5

Substitute into 2) to check if we did it correctly.

2(5) + 4(-5) + 10 = 0

10 -20 + 10 = 0

0=0

Good luck.

Paul

Feb 10, 2015 | Office Equipment & Supplies

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | SoftMath Algebrator - Algebra Homework...

this is a simple algebraic equation in which we have to calculate the value of 'x' using simple operations

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

step 1(pen the brackets) 7+9-5x=10-5x

step 2(add 7 and 9) 16-5x=10-5x

now as 5x gets cancelled both the sides of the equal sign therefore no solution of this algebraic equation exist.

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

step 1(pen the brackets) 7+9-5x=10-5x

step 2(add 7 and 9) 16-5x=10-5x

now as 5x gets cancelled both the sides of the equal sign therefore no solution of this algebraic equation exist.

Mar 24, 2011 | MathRescue Word Problems Of Algebra Lite

Hello this is Baris,

First off all when you say hypotinuse I think it is a 90 Degree triangle. Total of the angles off a triangle is 180 degree. So it is a simple algebra question.

Hypotinuse is 90

Right angle is R

Left angle is L

Total is 180

Equation is 90+ R +L = 180

We are looking for R, so lets arrange our equation.

R= 180 - 90 - L

R= 90 - L

This is our formula for excel

Lets put it in the excel

Type in A1 "hypotinuse "

Type in A2 "Left Angle"

Type in A3 "Right Angle"

Type in B1 90

Type in B2 "whatever the value given for the left angle"

Type in B3 =90-B2

As soon as you change the value in the cell B2, B3 will change by itself. Hope this helps :)

First off all when you say hypotinuse I think it is a 90 Degree triangle. Total of the angles off a triangle is 180 degree. So it is a simple algebra question.

Hypotinuse is 90

Right angle is R

Left angle is L

Total is 180

Equation is 90+ R +L = 180

We are looking for R, so lets arrange our equation.

R= 180 - 90 - L

R= 90 - L

This is our formula for excel

Lets put it in the excel

Type in A1 "hypotinuse "

Type in A2 "Left Angle"

Type in A3 "Right Angle"

Type in B1 90

Type in B2 "whatever the value given for the left angle"

Type in B3 =90-B2

As soon as you change the value in the cell B2, B3 will change by itself. Hope this helps :)

Sep 27, 2009 | Microsoft Excel for PC

Assuming M is the intersection of MN with AB, and N is the intersection of MN and AC:

Angle ACP = angle BCP (by definition)

Angle NCP = angle BCP (intersection of line with parallel lines produces equal angles)

Triangle CPN is isoceles (two equal angles), and line NP = CN

Same argument for line MP = BM

Therefore NP + MP (i.e, MN) = CN + BM

Angle ACP = angle BCP (by definition)

Angle NCP = angle BCP (intersection of line with parallel lines produces equal angles)

Triangle CPN is isoceles (two equal angles), and line NP = CN

Same argument for line MP = BM

Therefore NP + MP (i.e, MN) = CN + BM

Sep 08, 2009 | Canon Office Equipment & Supplies

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