Answer:
i can help with a littel bit of it
Step-by-step explanation:
1. answer
29±95‾‾‾√9
x
=
−
2
9
±
95
i
9
=−0.222222+1.08298
x
=
−
0.222222
+
1.08298
i
=−0.222222−1.08298
x
=
−
0.222222
−
1.08298
i
Find the Solution for:
92+4+11=0
9
x
2
+
4
x
+
11
=
0
using the Quadratic Formula where
a = 9, b = 4, and c = 11
=−±2−4‾‾‾‾‾‾‾‾√2
x
=
−
b
±
b
2
−
4
a
c
2
a
=−4±42−4(9)(11)‾‾‾‾‾‾‾‾‾‾‾‾√2(9)
x
=
−
4
±
4
2
−
4
(
9
)
(
11
)
2
(
9
)
=−4±16−396‾‾‾‾‾‾‾‾‾√18
x
=
−
4
±
16
−
396
18
=−4±−380‾‾‾‾‾√18
x
=
−
4
±
−
380
18
The discriminant 2−4<0
b
2
−
4
a
c
<
0
so, there are two complex roots.
Simplify the Radical:
=−4±295‾‾‾√18
x
=
−
4
±
2
95
i
18
=−418±295‾‾‾√18
x
=
−
4
18
±
2
95
i
18
Simplify fractions and/or signs:
=−29±95‾‾‾√9
x
=
−
2
9
±
95
i
9
which becomes
=−0.222222+1.08298
x
=
−
0.222222
+
1.08298
i
=−0.222222−1.08298
2. Answer:
=−38±895‾‾‾‾√40
x
=
−
3
8
±
895
i
40
=−0.375+0.747914
x
=
−
0.375
+
0.747914
i
=−0.375−0.747914
x
=
−
0.375
−
0.747914
i
3. Answer:
=0=−74
x
=
0
x
=
−
7
4
=0
x
=
0
=−1.75
Find the Solution for
82+14+0=0
8
x
2
+
14
x
+
0
=
0
using the Quadratic Formula where
a = 8, b = 14, and c = 0
=−±2−4‾‾‾‾‾‾‾‾√2
x
=
−
b
±
b
2
−
4
a
c
2
a
=−14±142−4(8)(0)‾‾‾‾‾‾‾‾‾‾‾‾√2(8)
x
=
−
14
±
14
2
−
4
(
8
)
(
0
)
2
(
8
)
=−14±196−0‾‾‾‾‾‾‾√16
x
=
−
14
±
196
−
0
16
=−14±196‾‾‾‾√16
x
=
−
14
±
196
16
The discriminant 2−4>0
b
2
−
4
a
c
>
0
so, there are two real roots.
Simplify the Radical:
=−14±1416
x
=
−
14
±
14
16
=016=−2816
x
=
0
16
x
=
−
28
16
=0=−74
x
=
0
x
=
−
7
4
which becomes
=0
x
=
0
=−1.75
i hope this helps
Answer:
Burgers: 12
French Fries: 34
Step-by-step explanation:
Answer:
Answer choice A
Step-by-step explanation:
This is the midpoint of both coordinates.
The coordinates of A will be (2P +M)/3
= (2(16, 14) +(1, 4))/3 = (33/3, 32/3) = (11, 32/3)
The appropriate choice is
(C) (11, 32/3)
_____
You will note that the coordinates of A are the weighted average of the coordinates of the end points. The weighting is the reverse of the ratio of the line segments. That is, the point adjacent to the shortest segment gets the highest weighting. (This is typical of the solution to "mixture" problems.)
507 peaches in each box.
He gives away 5 peaches.
If he sells 7 boxes he has 1019 peaches left.
Hope this helps.
