All categories

3 years ago

Use the process of completing the square to fill in the blanks. y = x2 + 8x + 25 y = (x + a0)2 + a1

1 answer:

Alex3 years ago

5 0

It would be y=(x+4)²+9.

To complete the square, we divide the value of b, the coefficient of x, by 2 and square it:

(8/2)² = 4² = 16

We would add 16 to this in order to have a perfect square; but we would also need to subtract 16 at the end to keep it equal. This would give us:

y=(x²+8x+16)+25-16
y = (x+4)² + 9

You might be interested in

How do you make. Two linear equations that are perpendicular to each other

s2008m [1.1K]

Answer:

In order to make two linear equations perpendicular to each other, we need to find the negative reciprocal of the slope.

In finding the negative reciprocal, we create a perfect 90° angle with the lines.

Example:

Given y = 2x + 3, find the perpendicular line.

2 -> -1/2

y = -1/2x + 3

8 0

3 years ago

Read 2 more answers

5.3 in.<br> 8 in.<br> 4 in.<br> 8 in.

trapecia [35]

Answer: 58.6

Step-by-step explanation:

4 x 8 = 32

4 x 5.3 = 21.2

4 x 2.7 = 10.8 / 2 = 5.4

32 + 21.2 + 5.4 = 58.6

4 0

3 years ago

Here is some information about 26 houses.a,b and c are all different numbers.Number of bedrooms:1,2,3,4,5.Number of houses:7,a,b

kkurt [141]

Answer:

$[a,b,c]=[4,2,5]$ or $[a,b,c]=[2,4,5]$

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Bedroom} & {1} & {2} & {3} & {4} & {5} \ \\ {Houses} & {7} & {a} & {b} & {c} & {8} \ \end{array}$

$n = 26$

$Median = 3.5$

Required

Find a, b and c

Median is calculated as:

$Median = \frac{n+1}{2}$

$Median = \frac{26+1}{2}$

$Median = \frac{27}{2}$

$Median = 13.5th$

This means that the median is the average of the 13th and 14th item.

$\begin{array}{cccccc}{Bedroom} & {1} & {2} & {3} & {4} & {5} \ \\ {Houses} & {7} & {a} & {b} & {c} & {8} \ \end{array}$

Since the median is $Median = 3.5$ (average of 3 and 4),

and:

$3 \to b \to 13$

$4 \to c \to 14$

This implies that c begins at 14

So, the number of houses bedrooms 4 and 5 is:

$c + 8 = 26 - 14 +1$

$c + 8 = 13$

$c = 5$

Also

$7 + a + b + c + 8 = 26$ ---- sum of total frequency

$a + b + c = 26 - 7 - 8$

$a + b + c = 11$

Substitute $c = 5$

$a + b + 5 = 11$

$a + b = 11 - 5$

$a + b = 6$

Since a, b and c are different, then

a and b cannot be 5 because $c = 5$

a and b cannot be 3 because $a = b =3$

$b \ne 0$ because b ends at 13

So, possible sets are:

$[a,b]=[2,4]$

$[a,b]=[4,2]$

Include c, we have:

$[a,b,c]=[4,2,5]$ or $[a,b,c]=[2,4,5]$

5 0

3 years ago

The histograms below show the mass (in kilograms) of the offensive linemen on the University of Mount Union and the University o

Sedbober [7]

Answer:

Option C is correct.

None of the options is true.

Step-by-step explanation:

The histograms for the question is missing, so, it was obtained online and is attached to this solution of the question.

To check which statements are true, we first find the average using the midpoint technique.

The midpoint for each of the ranges is presented below

110 to 120 = 115

120 to 130 = 125

130 to 140 = 135

140 to 150 = 145

150 to 160 = 155

Sum of weights for Alabama linemen using the midpoint technique

= (125×4) + (135×8) + (145×4) + (155×2)

= 2470 kg

Average = (2470) ÷ (4+8+4+2)

= 137.22 kg

Sum of weights for Mount Union linemen

= (115×5) + (125×4) + (135×6) + (145×2)

= 2,175 kg

Average = (2,175) ÷ (5+4+6+2)

= 127.94 kg

So, it is evident that the average weight of Mount Union linemen isn't more than that of Alabama linemen.

Statement 2

All of the Alabama linemen have more mass than all of the Mount Union linemen

This is evidently not true because from the histogram, some linemen from Alabama have masses between 120 and 130, 130 and 140 & there are linemen from Mount Union that have masses between 130 and 140 & 140 and 150, so, it isn't right to say that all of the Alabama linemen have more mass than all of the Mount Union linemen.

So, none of the options is correct, option C.

Hope this Helps!!!

3 0

3 years ago

Which of the following best defines an output

Finger [1]

The answer is the 3rd option im pretty sure :)

4 0

3 years ago