All categories

3 years ago

What is the difference of the two polynomials? (9x2 + 8x) – (2x2 + 3x)

Mathematics

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:

(9x² + 8x) - (2x² + 3x)

= 7x² + 5x

Step-by-step explanation:

(9x² + 8x) - (2x² + 3x)

open the bracket

= 9x² + 8x - 2x² - 3x

collect like terms

= 9x² - 2x² + 8x - 3x

= 7x² + 5x

damaskus [11]3 years ago

7 0

(9x^2+8x)-(2x^2+3x)
9x^2+8x-2x^2-3x
7x^2+5x

You might be interested in

I am an odd number. When I am tripled, I am one more than 20. Which number sentence identifies the odd number, n?

Andreyy89

Is A. Because 3n= 20+1

4 0

3 years ago

A rectangle has a perimeter of 30 inches. One Side measures 5 inches.What are the measurements of the other three sides?

cluponka [151]

Let's say the two on the side are 5 inches, then you're left with 20 inches left. 20/2 sides = 10 inches per side. So the measurements are 5 in, 5 in, 10 in, 10 in.

5 0

3 years ago

Read 2 more answers

If you where to subtract 5 from my number and then divide the difference by 8 you would get 6. What is my number. Write an equat

Brut [27]

do it in reverse so 6 x 8 is 48 + 5 is 53.

3 0

4 years ago

Which quadratic function in vertex form can be represented by the graph that has a vertex at (3, -7) and passes through the poin

yarga [219]

$~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=3\\ k=-7 \end{cases}\implies y=a(x-3)^2-7\qquad \textit{we also know that} \begin{cases} x=1\\ y=-10 \end{cases} \\\\\\ -10=a(1-3)^2-7\implies -3=a(-2)^2\implies -3=4a\implies -\cfrac{3}{4}=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-\cfrac{3}{4}(x-3)^2-7~\hfill$

5 0

2 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0

3 years ago