Answer:
I have listed all the dashes according to numbers so we have total of 11 dashes there.
I'll be answering them according to the assumed number of dash.
Step-by-step explanation:
- 13
- 8/13
- 13/21
- 0.625000
- 0.615385
- 0.619047
- 8/5
- 13/8
- 1.666667
- 1.600000
- 1.625000
Hope you understand it
X actually equals the cube root of 6
![{x}^{3} - 6 = 0 \\ {x}^{3} = 6 \\ x = \sqrt[3]{6}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B3%7D%20%20%20-%206%20%3D%200%20%5C%5C%20%20%7Bx%7D%5E%7B3%7D%20%20%3D%206%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B6%7D%20)
If you download Photomath and scan your question it will give u the answer;)
First picture)
I: 5x+2y=-4
II: -3x+2y=12
add I+(-1*II):
5x+2y-(-3x+2y)=-4-12
8x=-16
x=-2
insert x=-2 into I:
5*(-2)+2y=-4
-10+2y=-4
2y=6
y=3
(-2,3)
question 6)
I: totalcost=115=3*childs+5*adults
II: 33=adults+childs
33-adults=childs
insert childs into I:
115=3*(33-adults)+5*adults
115=99-3*adults+5*adults
16=2*adults
8=adults
insert adults into II:
33-8=childs
25=childs
so it's the last option
question 7)
a) y<6 and y>2 can also be written as 2<y<6, so solution 3 exist for example
b) y>6 and y>2 can also be written as 2<6<y, so solution 7 exist for example
c) y<6 and y<2 inverse of b: y<2<6, so for example 1
d) y>6 and y<2: y<2<6<y, this is impossible as y can be only either bigger or smaller than 2 or 6
so it's the last option
question 8)
I: x+y=12
II: x-y=6
subtract: I-II:
x+y-(x-y)=12-6
2y=6
y=3
insert y into I:
x+3=12
x=9
(9,3)
question 9)
I: x+y=6
II: x=y+5
if you take the x=y+5 definition of II and substitute it into I:
(y+5)+y=6
which is the second option :)
