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Given that

\(2(-a-b-c)^3-27/8=9(-a-b-c)(ab+bc+ca\)

and

\(4(ab+bc+ca)+6abc=-9(a+b+c)+9(ab+bc+ca)-9abc+9 \)

Find a+b+c.

Hello Guest!

\( 2(-a-b-c)^3-27/8=9(-a-b-c)(ab+bc+ca)\)

wolframalpha.com:

\(a\approx -1.11502\\ b\approx 0.162679\\ c\approx -1.75387\\ \color{blue}a+b+c\approx -2.706211\)

\(4(ab+bc+ca)+6abc=-9(a+b+c)+9(ab+bc+ca)-9abc+9 \)

wolframalpha.com

Integer solutions:

\(a\in \{0,0,0,0,0\}\\ b\in \{0,1,2,3,9\}\\ c\in \{1,0,9,3,2\}\)

\((a+b+c)\in\{1,1,11,6,11\}\)

  !

A line passes through (-6,2) and (-2,7), making a triangle in the second quadrant.  Find the number of square units in the area of the triangle.

Hello Guest!

\(y=m(x-x_1)+y_1\)

\(m=\frac{7-2}{-2+6}=\frac{5}{4}\)

\(y=\frac{5}{4}(x+6)+2\)

\({\color{blue}y=\frac{5}{4}x+9.5}\\ x_0=-7.6\\ y_0=9.5\)

\(A_{◿}=\frac{1}{2}\cdot 7.6\cdot 9.5\)

\(A_{◿}=36.1\ square\ units\)

  !

The side across from the 60° angle is √3 times the side across from the 30° angle,

and the side across from the 30° angle is half the hypotenuse.  So...

a   =   8/2 * √3   =   4√3

b   =   a/2 * √3   =   4√3 / 2  *  √3   =   2√3 * √3   =   2 * 3   =   6

c   =   b/2 * √3   =   6/2 * √3   =   3√3

d   =   c/2   =   3√3 / 2

(where a, b, c, and d are all in cm)

math = physics 

a single line is not enough to make a triangle.

The domain of g is [-1,3], so the witdh of the interval is 4.

Wow.....really detailed long question.....most people know what a deck o' cards contains....   NEway:

4 aces and 12 +12 more red cards ....     28 out of 52

Kelly's art teacher asked him to draw a rectangle with a length of 6 inches and a width of 10 inches. After Kelly does so, his teacher changes her mind and asks him to draw a proportional rectangle that has a length of 9 inches. How many inches wide is the new rectangle?

Please don't chnage your question after an answer has been posted.....just post a new question .....o/w it is very confusing

Teacher changed the dimension from length of 6    to length of  9

   length changed by a factor of 1.5     (because   6 x 1.5 = 9)

      width should chnage by the same proportion   10 * 1.5   =   15 inches

xxxxxxxxxxxxxx

Hello Guest.

For a), It is could be congruent, since AAA is not a congruence, but a similarity.

For b), it is not congruent, since the triangle is not an iscoceles triangle and we have te values of two not-corresponding sides.

I hope this helped! 

Hello Guest.

"Andrew is building a scale model of an elephant. He determines that a real elephant is 10 feet tall and 14 feet long. If Andrew wants his model to be 8 inches tall, how many inches long should his model be?"

SOLUTION:

We can use ratios to help us solve this problem here.

We already know the ratio for height, which is 10:8. Now, we just have to set the width of his drawing to $x$ and write out an equivalent ratio for that: \[\frac{10}{8}=\frac{14}{x}.\] Simplifying and cross multiplying, we get $\boxed{x=\frac{56}{5}}$

I hope this helped! 

Thanks!

Hello EvenWei.

This seems to be a repeat of the question given here: https://web2.0calc.com/questions/help-plz_7543

If my answer was incorrect, please let me know 

Thanks!

Something's wrong here! The question calls for the area of triangle ABM, but point M is on line segment AB

It's more likely the triangle ABN   ... or something else. (?!?)

Answer: 3

\(A(3,4) \) reflected over the X-axis is \(B(-3,4)\).

\(B(-3,4)\) reflected over y=-x is \((-y,-x)\) which means that C is \(C(-4,3)\)

If we create an imaginary rectangle, the area of that is 7. Subtract the corners, and we get \(7-\frac{1}{2}-\frac{7}{2}=3\)

I think you just type it in via your keyboard

example

5!   then =

find side length of triangle

1/2 b h = 8 sqrt3

bh = 16 sqrt3

b  * b sin60 = 16 sqrt3

b^2  sqrt3/2 = 16 sqrt 3

b^2 = 32

b = sqrt 32

so perimeter is   3 sqrt 32      square side is then  3/4 sqrt 32  

    diagonal^2  is     2 (3/4 sqrt32)^2  = 18/16 *32              d = 6 in