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

write the negative exponents which are numerators as denomenators and vice versa

so A^-3b = 1/A^3b etc

result (b^[31/3] a^9 )/ (A^3b b^6)

25.March.2015

so A^-3b = 1/A^3b etc

result (b^[31/3] a^9 )/ (A^3b b^6)

25.March.2015

Mar 25, 2015 | MathAid Algebra II

Assuming the 'standard form' is "slope-intercept", calculate the slope from the equation m = __y2-y1__ =__ 5 - 1__ = __ 4__ = -2

x2-x1 4 - 6 -2

The intercept can be found by substituting either of the two points into the equation y = mx + b

5 = (-2)4 + b

5 = (-8) + b

13 = b

(OR, using the other point, y = mx + b

1 = (-2)6 + b

1 = (-12) + b

13 = b )

Then expressing in general:

**y = (-2) x + 13**

x2-x1 4 - 6 -2

The intercept can be found by substituting either of the two points into the equation y = mx + b

5 = (-2)4 + b

5 = (-8) + b

13 = b

(OR, using the other point, y = mx + b

1 = (-2)6 + b

1 = (-12) + b

13 = b )

Then expressing in general:

Oct 10, 2014 | Computers & Internet

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...

A pyramid has 5 sides including the base. If the base is a rectangle L long and W wide, its area is L*W.

The other sides are triangles. If the height of the pyramid is H, the triangles that are L wide at the base will have a height of the square root of half of W squared plus H squared, or SQRT((.5*W)^2+H^2) and the two that are W wide at the base will have a height of SQRT((.5*L)^2+H^2). The area of a triangle is .5*height*base, so the total surface two triangles with the same height and base is 2*.5*height*base = height*base.

So, the total surface area of a pyramid L long by W wide by H tall would be:

L*W (the area of the base)

+ L*SQRT((.5*W)^2+H^2) (the area of the 2 triangles with base L)

+ W*SQRT((.5*L)^2+H^2) (the area of the 2 triangles with base W)

Hope this is helpful!

The other sides are triangles. If the height of the pyramid is H, the triangles that are L wide at the base will have a height of the square root of half of W squared plus H squared, or SQRT((.5*W)^2+H^2) and the two that are W wide at the base will have a height of SQRT((.5*L)^2+H^2). The area of a triangle is .5*height*base, so the total surface two triangles with the same height and base is 2*.5*height*base = height*base.

So, the total surface area of a pyramid L long by W wide by H tall would be:

L*W (the area of the base)

+ L*SQRT((.5*W)^2+H^2) (the area of the 2 triangles with base L)

+ W*SQRT((.5*L)^2+H^2) (the area of the 2 triangles with base W)

Hope this is helpful!

May 23, 2014 | Encore Math Advantage Algebra II and...

-1 (negative one).

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

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

Apr 30, 2013 | 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

I'm assumeing the problem is 2x^2 - 19x+22 = 0, solve for x.

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.

Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =

(19 +/- squareroot(185))/4 =

8.150367 or 1.349632

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.

Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =

(19 +/- squareroot(185))/4 =

8.150367 or 1.349632

Feb 01, 2011 | SoftMath Algebrator - Algebra Homework...

Hi joanmae jmeh;

You can add these equations just like you would numbers

3x + 2y = 7

5x - 2y = 1

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

8x =8

x=1

Now plug x = 1 back into either equation and you can solve for y

Y = 2

When you add the 2 equations together the +2y and the -2y cancel

out

Hope this helps Loringh Please leave a rating for me Thks

You can add these equations just like you would numbers

3x + 2y = 7

5x - 2y = 1

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

8x =8

x=1

Now plug x = 1 back into either equation and you can solve for y

Y = 2

When you add the 2 equations together the +2y and the -2y cancel

out

Hope this helps Loringh Please leave a rating for me Thks

Dec 01, 2008 | Bagatrix Algebra Solved! 2005 (105101) for...

Hi rowanwah

The sine of an angle is only applicable is a right triangle. If you just want a number, ie, the actual value of the sine 15 degrees you can look it up on Google. Do a search for "sine and cosine functions"

If you want the mathematical description of the sine of an angle it is described as follows

In a triangle ABC, there are 3 angles angle A, angle B and angle C. There are also 3 sides, Side AB, Side AC and side BC. The sine of angle A is equal to the side opposite Angle A divided by the Hypotenuse (the longest side opposite the right angle)

The Cosine of angle A is equal to the side adjacent to Angle A divided by the hypotenuse

Hope this helps Loringh PS Please leave a rating for me Thanks

The sine of an angle is only applicable is a right triangle. If you just want a number, ie, the actual value of the sine 15 degrees you can look it up on Google. Do a search for "sine and cosine functions"

If you want the mathematical description of the sine of an angle it is described as follows

In a triangle ABC, there are 3 angles angle A, angle B and angle C. There are also 3 sides, Side AB, Side AC and side BC. The sine of angle A is equal to the side opposite Angle A divided by the Hypotenuse (the longest side opposite the right angle)

The Cosine of angle A is equal to the side adjacent to Angle A divided by the hypotenuse

Hope this helps Loringh PS Please leave a rating for me Thanks

Nov 15, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

Divide the first equation by 2, the second by 6 (i.e. the number that comes before the x)

Oct 04, 2008 | SoftMath Algebrator - Algebra Homework...

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