11. You’ve done it correctly
12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2
13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution
14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath
15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4
Hope this helped :)
Answer:
the answer would be 1060
Step-by-step explanation:
all you have to do is multiply 260 by four and then you have your answer
Maximum number of packets = 8
16 ÷ 8 = 2
There are 2 pencils in each packet.
24÷8 = 3
There are 3 erasers in each bag.
Rosy can have a maximum of 8 packets. Each packet has 2 pencils and 3 erasers.
Answer:
(x−3)(x−5)
Step-by-step explanation:
Let's factor x2−8x+15
x2−8x+15
The middle number is -8 and the last number is 15.
Factoring means we want something like
(x+_)(x+_)
Which numbers go in the blanks?
We need two numbers that...
Add together to get -8
Multiply together to get 15
Can you think of the two numbers?
Try -3 and -5:
-3+-5 = -8
-3*-5 = 15
Fill in the blanks in
(x+_)(x+_)
with -3 and -5 to get...
(x-3)(x-5)
Answer:
Option 3 (C)
Step-by-step explanation:
Given the scores of the students: 3, 4, 4, 5, 5, 6, 8, 10, 10, 12, 15, 18
To find the box plot that represents the data set, we need to find:
Min, Q1, median, Q3, and Max, which are all represented on a box plot.
Min = 3
Max = 18
Median = the average of the 6th and 7th data value = (6+8)/2 = 14/2 = 7
Q1 = the average of the 3rd and 4th data values = (4+5)/2 = 4.5
Q3 = the average of the 9th and 10th data values = (10+12)/2 = 11
The box plot that closely represented all these values is the box plot in option 3.
