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3 years ago

Which of the following is a perfect square trinomial?

2 answers:

6 0

Your answer is D. 16x² - 56xy + 49y².

A perfect square trinomial is the result of a squared binomial, like (a + b)². Using this example, the perfect square trinomial would be a² + 2ab + b², as that is what you get when you expand the brackets.

Therefore, to determine which of these is a perfect square trinomial, we have to see if it can be factorised into the form (a + b)².

I did this by first square rooting the 16x² and 49y² to get 4x and 7y as our two terms in the brackets. We automatically know the answer isn't A or B as you cannot have a negative square number.

Now that we know the brackets are (4x + 7y)², we can expand to find out what the middle term is, so:

(4x + 7y)(4x + 7y)
= 16x² + (7y × 4x) + (7y × 4x) + 49y²
= 16x² + 28xy + 28xy + 49y²
= 16x² + 56xy + 49y².

So we know that the middle number is 56xy. Now we assumed that it was (4x + 7y)², but the same 16x² and 49y² can also be formed by (4x - 7y)², and expanding this bracket turns the +56xy into -56xy, forming option D, 16x² - 56xy + 49y².

I hope this helps!

AVprozaik [17]3 years ago

6 0

Answer:

It would have to be the last choice

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Please help! Brainliest

Mama L [17]

1.301

Explanation:

Given:

$\log_a5 \approx 0.699$ and $\log_a2 \approx 0.301$

Note that from the properties of logarithms,

$\log_a20 = \log_a(5×4)$

$\:\:\:\:\:\:\:\:\:\:= \log_a(5×2^2)$

$\:\:\:\:\:\:\:\:\:\:=\log_a5 + \log_a(2)^2$

$\:\:\:\:\:\:\:\:\:\:=\log_a5 + 2\log_a2$

$\:\:\:\:\:\:\:\:\:\:\approx 0.699 + 2(0.301)$

$\:\:\:\:\:\:\:\:\:\:= 1.301$

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3 years ago

Please answer asap <br> What is the value of x in the equation 3(2x + 4) = −6?

nydimaria [60]

First divide 3 both side
Than move 4 to another side
Than divide by 2
$\frac{3(2x + 4)}{3} = \frac{ - 6}{3 } \\ 2x + 4 = - 2 \\ 2x = - 2 - 4 \\ 2x = - 6 \\ \frac{2x}{2} = \frac{ - 6}{2} \\ x = - 3$

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3 years ago

What is the reciporcal of 3 and 1/3

aev [14]

Answer:

your answer is 0.3

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6 0

3 years ago

Read 2 more answers

We are told that the data is representative of the two populations (U.S. males aged 20-29 years and U.S. males aged 75+ years),

tiny-mole [99]

Answer:

The data provide strong evidence that young men weigh more on average than old men in the U.S

Step-by-step explanation:

Given :

The null hypothesis ; H0 : μ1 = μ2

The alternative hypothesis ; H1 : μ1 > μ2

T score = 5.3 ; Pvalue = < 0.0001

The decision region :

If Pvalue < α ; We reject the Null

If Pvalue > α ; We fail to reject the Null

When the α - level isn't stated, we usually assume a α - level of 5%

However, even at lower alpha level of 1% = 0.01 ;

The Pvalue < α

Hence, we can conclude that there is significant evidence that there is difference in the mean weight of young men and old men in the U.S

3 0

3 years ago

The mean maximum aerobic power (VO2MAX) score for women ages 20 to 29 is 36 ml/min/kg with a standard deviation of 7 ml/min/kg.

Scrat [10]

Answer:

0.099 is the probability of a woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 36 ml/min/kg

Standard Deviation, σ = 7 ml/min/kg

We assume that the distribution of aerobic power is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.)

P(x > 45)

$P( x > 45) = P( z > \displaystyle\frac{45 - 36}{7}) = P(z > 1.285)$

$= 1 - P(z \leq 1.285)$

Calculation the value from standard normal z table, we have,

$P(x > 45) = 1 - 0.901 =0.099 = 9.9\%$

0.099 is the probability of a woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.

4 0

3 years ago