All categories

2 years ago

Find the product of (3x + 7y)(3x − 7y)

Mathematics

2 answers:

snow_lady [41]2 years ago

8 0

Answer: 9x^2 - 49y^2

Step-by-step explanation:

(3x + 7y)(3x − 7y)

9x^2 -21xy + 21xy - 49y^2

9x^2 - 49y^2

nalin [4]2 years ago

7 0

Answer:

9x^2 - 49y^2

Step-by-step explanation:

You might be interested in

1. x/6 + 4 = 15

Anestetic [448]

1. x/6 = 15 - 4
x = 11 × 6
x = 66

2. 2/7x = 3/14 - 1/7
2/7x = 3 - 2/14
2 = 7x/14
x = 4

3. 4+10 = x/2
14 × 2 = x
x = 28

8 0

2 years ago

Solve the inequality 3x - 5 < 4.

lbvjy [14]

6 0

2 years ago

Read 2 more answers

Maya’s chocolate is 59% cocoa. If the weight of the Chocolate bar is 93 grams, how many grams cocoa does it contain? Round your

Natasha2012 [34]

Answer:

54.9 because it is the answer

7 0

2 years ago

Read 2 more answers

If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

zmey [24]

Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

4 0

2 years ago

Read 2 more answers

Suppose that a standardized biology exam has a mean score of 80% correct, with a standard deviation of 3. The school administrat

NemiM [27]

Answer:

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.

This means that $n = 65, \pi = 0.76$

99% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 - 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.6236$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.8964$

0.6236*100 = 62.36%

0.8964*100 = 89.64%

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

6 0

2 years ago