PLEASE HELP ME ( MULTIPLE CHOICE) WILL GIVE BRAINLIST

I just need an explanation on why I was wrong and why the correct answer is correct

Alja [10]

Answer:

B 1

Step-by-step explanation:

Since the divisor is in the form of x - c, use what is called Synthetic Division. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

2| -2 1 5 0 4 1

↓ -4 -6 -2 -4 0

_________________

-2 -3 -1 -2 0 1→ -2x⁴ - 3x³ - x² -2x + [x - 2]⁻¹

You start by placing the c in the top left corner, then list all the coefficients of your dividend [-2x⁵ + x⁴ + 5x³ + 4x + 1]. You bring down the original term closest to c then begin your multiplication. Now depending on what symbol your result is, tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that -2 in your quotient can be a -2x⁴, and the -3 [x³] follows right behind it, then 1 [-x²], -2[x], and finally, [1\x - 2] (remainder is 1, so set it over your denominator, which is the divisor), giving you the other factor of -2x⁴ - 3x³ - x² -2x + [x - 2]⁻¹.

I am joyous to assist you anytime.

** $(x - 2)^{-1} = \frac{1}{x - 2}$

5 0

3 years ago

A kangaroo hops 2 kilometers in 3 minutes. At this rate, how far does the kangaroo travel in 2 minutes? *

Dmitrij [34]

Answer:

At a speed of 2 km every 3 minutes, in 2 minutes the kangaroo would travel 1,333 km.

Step-by-step explanation:

If a kangaroo hops 2 kilometers in 3 minutes, to calculate how much distance this animal hops in 2 minutes, the following calculation should be performed:

2 km / 3 min = 0.666 km per minute (2/3 x 2)

0.666 km per minute x 2 minutes = 1.333 kilometers in 2 minutes (0.666 x 2)

Therefore, at a speed of 2 km every 3 minutes, in 2 minutes the kangaroo would hop 1,333 km.

5 0

2 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$