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3 years ago

The polynomial below is a perfect square trinomial of the form A2 - 2AB + B2.

Mathematics

2 answers:

sertanlavr [38]3 years ago

8 0

Yes the statement is true.

irakobra [83]3 years ago

5 0

Answer:

True

Step-by-step explanation:

Given: $9x^2-30x+25$

We have to write the given polynomial in perfect square nominal.

$A^2-2AB+B^2$

We can factor it like

$A^2-2AB+B^2=(A-B)^2$

It is perfect square trinomial.

$9x^2-30x+25$

We have to make perfect square first and last term.

$9x^2\rightarrow (3x)^2\rightarrow A$

$25\rightarrow 5^2\rightarrow B$

$9x^2-30x+25\Rightarrow (3x)^2-2\cdot 3x\cdot 5+5^2$

Hence, The given polynomial is in perfect square trinomial.

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

Which of the following fractions can be represented by a terminating decimal?

Karo-lina-s [1.5K]

Answer:

B) 13/8

Step-by-step explanation:

First we need to understand what a terminating decimal is: It is a decimal that has a finite number of decimal values that are not zero, meaning that its decimal values end at some point

Let's go through our possible answers with trial and error:

A) 8/9

8/9 = 0.888888888888...9 (Incorrect)

B) 13/8

13/8 = 1.625 (Correct)

C) 4/3

4/3 = 1.3333333333...4 (Incorrect)

D) 6/11

6/11 = 0.545454545454...54 (Incorrect

13/8 is our answer because in decimal form it is a terminating decimal.

8 0

2 years ago

You run one lap around a mile track every eight minutes you're friend runs around the same track every 10 minutes. You both star

labwork [276]

10 , 20 , 30 , 40 , 50
8, 16 , 24, 32, 40

40 minutes

4 0

3 years ago

K+ 6 ≥ 19, if k = 11

WINSTONCH [101]

Answer:

Then it would be the other way around 19 ≥ k + 6

Wait for more responses if needed.

4 0

3 years ago

Read 2 more answers

Function g is a transformation of function f.

bulgar [2K]

The equation of function g(x) in terms of f(x) is g(x) = -3[f(x)].

<h3>What is an equation?</h3>

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

Given:

genera form of an exponential function is y=aeᵇˣ

Now, equation for f(x) is

f(x) = $e^{(log \;2)x} -2$

Similarly, graph for g(x) is

g(x) = $-3e^{(log \;2)x} +6$

Comparing the two function a relation can be establish

g(x) = -3[f(x)]

Learn more about Equation here:

brainly.com/question/2263981

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4 0

2 years ago