All categories

2 years ago

2x(3a-2x) + a(2x-a)-3x(a+x)

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

5 0

6xa - 4x(^2) + 2xa - a (^2) - 3ax - 3x (^2)
5 xa - 7 x(^2) - a (^2)

You might be interested in

The state tax on clothing is 5.50 percent. what is the cost of 25$ shirt, including tax?

Natalija [7]

5.50% = 0.0550

25 + 0.0550(25) = 25 + 1.38 = 26.38

3 0

3 years ago

What is the median if the following numbers:10,10,12,14,22,13, and 24

mina [271]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Simplify 24 – 6x – 18 + 5x 3 – 3x –x + 6 2x – 3 –4x – 17

Svetach [21]

= −x+6

= −3x+3

= −x+6
= 2x−3

= −4x−17

6 0

3 years ago

Find the values of the six trigonometric functions of 0

Leni [432]

Step-by-step explanation:

sin theta = 12/13

cos theta = 5/13

tan theta = 12/5

cosec theta = 13/12

sec theta = 13/5

cot theta = 5/12

8 0

3 years ago

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two x-coordinates.

To find the y-coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

6 0

3 years ago