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3 years ago

Finding the roots by completing the square

1 answer:

6 0

For y=x^2-6x-11

complete the square

so roots
set y=0
0=x^2-6x-11

group x terms
0=(x^2-6x)-11
take 1/2 of linear coefient (-6) and square it
-6/2=-3, (-3)^2=9
add positive and neative to inside of parenthasees
0=(x^2-6x+9-9)-11
complete square
0=((x-3)^2-9)-11
expand
0=(x-3)^2-9-11
0=(x-3)^2-20
add 20 to both sides
20=(x-3)^2
sqrt both sides
remember positive and negative roots

+/-2√5=x-3
add 3 to both sides
3+/-2√5=x

so
x=3+2√5 and 3-2√5

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The aquarium charges an entrance fee of $5 per child and $12 per adult. On one day, 100 people visited the aquarium. They collec

igor_vitrenko [27]

Answer:

We set up 2 equations

A) C + A = 100

B) 5C + 12A = 780

We multiply A by -5

A) -5C -5A = -500 then we add B

B) 5C + 12A = 780

7A = 280

Number of Adults = 40

5C = 780 - 40*12

5C = 780 -480

5C = 300

Number of Children = 60

Step-by-step explanation:

6 0

2 years ago

Mark wants to improve his free throw percentage in basketball He has shot 1,250 free throws over the last 20 days. He shot eithe

Sedaia [141]

Let's use a for number of days when he shot 50 shots and b for number of days when he shot 100 shots.
We have:
a + b = 20

We also know that he shot total of 1250 shots:
50a + 100 b = 1250

We have two equations. We can solve them for a and b. Let's rearange first equation for a:
a= 20 - b
We insert this into second equation:
50 * (20 - b ) + 100b = 1250
1000 - 50b + 100b = 1250
50b = 250
b = 5
a = 20 - 5
a = 15

Mark shot 100 shots on 5 days.

4 0

3 years ago

Juanita is making necklaces to give as presents. She plans to put 15 beads on each necklace. Beadsare sold in packages of 20. Wh

PtichkaEL [24]

3 packages because you'll have 15 beads left over. which would complete the use of the packages

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3 years ago

Read 2 more answers

Question 14

Reil [10]

Answer:

The system of equations has a one unique solution

Step-by-step explanation:

To quickly determine the number of solutions of a linear system of equations, we need to express each of the equations in slope-intercept form, so we can compare their slopes, and decide:

1) if they intersect at a unique point (when the slopes are different) thus giving a one solution, or

2) if the slopes have the exact same value giving parallel lines (with no intersections, and the y-intercept is different so there is no solution), or

3) if there is an infinite number of solutions (both lines are exactly the same, that is same slope and same y-intercept)

So we write them in slope -intercept form:

First equation:

$6x+y=-1\\y=-6x-1$

second equation:

$-6x-4y=4\\-6x=4y+4\\-6x-4=4y\\y=-\frac{3}{2} x-1$

So we see that their slopes are different (for the first one slope = -6, and for the second one slope= -3/2) and then the lines must intercept in a one unique point. Therefore the system of equations has a one unique solution.

6 0

3 years ago

Remove the parentheses and simplify 3a^2-6ab+3b^2-5(-6a^2+7ab-7b^2)

Oksi-84 [34.3K]

$3a^2-6ab+3b^2-5(-6a^2+7ab-7b^2)=\\\\3a^2-6ab+3b^2+30a^2-35ab+35b^2=\\\\33a^2-41ab+38b^2$

4 0

3 years ago