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3 years ago

Multiply (x-1)(x³+4x²+4x-1) and combine like terms. Explain how you reached your answer.

Mathematics

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

x⁴+3x³ -5x +1

Step-by-step explanation:

To multiply this we are going to multiply each term of the binomial by each term of the polynomial.

$(x-1)(x^{3}+4x^{2} +4x-1)\\=x^4+4x^3+4x^2-x -x^3-4x^2-4x+1$

Now we are going to combine like terms (this means we are going to sum up terms with the same exponent)

$x^4+4x^3+4x^2-x -x^3-4x^2-4x+1 = x^4 +(4x^3-x^3)+(4x^2-4x^2)+(-x-4x)+1\\=x^4+3x^3-5x+1$

Thus, the answer is x⁴+3x³ -5x +1

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If dy/dx = y cos x and y = 3 when x = 0, then y =....? Here are the answer choices

creativ13 [48]

Dy/dx = y cos x
y = Ae^sin(x)
y(0) = Ae^sin(0) = A = 3
A = 3

Therefore, y = 3e^(sinx)

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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:

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

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3 years ago

This hanger is in balance. There are two labeled weights of 5 grams and one of 33 grams. The four circles each have the same wei

kvv77 [185]

Answer:

Step-by-step explanation:

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