All categories

2 years ago

Factor completely 4n^2-28n+49

Mathematics

2 answers:

svetlana [45]2 years ago

6 0

Answer:

Then it would be 4n squared + 21

Step-by-step explanation:

topjm [15]2 years ago

6 0

Answer:

(2n - 7)^2

Step-by-step explanation:

both 4n^2 and 49 are perfect squares so take their square roots and get 2n and 7

You might be interested in

You have to work on Distributive property 14(5+4)

lukranit [14]

126

explanation: solved it

3 0

3 years ago

A square is folded into half to form a rectangle. the perimeter of the rectangle is 36cm. calculate the area of a square

zhannawk [14.2K]

Answer:

Let the measure of each side of the square be " x " cm.

then ,

area of square = x²

After we folded the square into a rectangle ,

length of rectangle = x

breadth of rectangle = x/2

perimeter of rectangle = 36 cm

$x + \frac{x}{2} + x + \frac{x}{2} = 36 \\ \\ \frac{2x + x + 2x + x}{2} = 36 \\ \\ \frac{6x}{2} = 36 \\ \\ 6x = 36 \times 2 \\ \\ 6x = 72 \\ \\ 6x = \cancel{\frac{72}{6} } = 12cm$

hence ,

side of square = 12 cm

therefore ,

area of square = 12 × 12 = 144 cm²

hope helpful~

8 0

2 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.