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3 years ago

Polynomial division help PLEASE!

Mathematics

1 answer:

inna [77]3 years ago

4 0

<h2>Answer:</h2>

sorry i don`t know. i will try to do it, give me a second and see if i can do it.

Step-by-step explanation:

You might be interested in

The ratio of water and milk in a mixture is 2:3, what percent of water is mixed in the mixture?

elena-s [515]

40% water 60% milk

reason:
if you make then go to 100 by multiplying 2:3 by 2 you get 4:6 and then add a 0 at the end and there’s your answer

5 0

2 years ago

Read 2 more answers

Five adult tickets and three child tickets for a movie costs £55. The cost of buying two adult tickets and three child tickets i

Allisa [31]

Answer:

adult £8
child £5

Step-by-step explanation:

If you look at the numbers you are given, you see that the first purchase has 3 more adult tickets than the second purchase, and its cost is £24 more. This means an adult ticket costs £24/3 = £8.

Two adult tickets will cost 2×£8 = £16, so three child tickets cost ...

£31 -16 = £15

Each child ticket is then £15/3 = £5.

An adult ticket costs £8; a child ticket costs £5.

7 0

3 years ago

Factor completely:

padilas [110]

Answer:

d) 2x²(6x³ + 3x + 4)

Step-by-step explanation:

In order to factor the given polynomial, you need to find the greatest common factor of all three terms: $12x^{5}+6x^{3}+8x^{2}$

The greatest common factor is '2' and the variable is x². If we factor out 2x² from each term, we get:

2x²(6x³ + 3x + 4)

8 0

2 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

2 years ago

A study is designed to examine the effect of doing synchronized movements (such as marching in step or doing synchronized dance

Vlada [557]

Answer:

The complete question in the graphic

After synchronized exercise, it was observed that the average proximity rate has increased, which shows that the mean of the variable is positive and there is high synchronization in the groups.

If groups with low synchronization and high effort are compared, the result of the average will be positive, so the average proximity rate will also increase after the synchronization exercise is performed.

On the contrary, if we observe the groups that have low synchronization and low effort, the mean of the variable is negative, thus the average proximity rate after performing the exercise will decrease.

(C)

considering that the total degrees of freedom is 259, it will show 260 students in the analysis.

With a significance level of 5% or 0.05, the P value is less than the significance level, which leads to differentiating the mean of the scores before and after performing the combinations.

With a significance level of 1% or 0.01, the p-value is higher than the significance level. With this data we can conclude that there is no difference between the results before and after the synchronized exercise

3 0

3 years ago