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____ [38]
3 years ago
9

Given the function f(x) = −2x + 8, find x if f(x) = 14 ...?

Mathematics
2 answers:
lisov135 [29]3 years ago
8 0
Solution:

14 = -2x +8
6 = -2x
3 = -x

x = -3
Over [174]3 years ago
5 0
14 = - 2x + 8
2x = - 6
x = - 3
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2 years ago
Lexi operates an orange juice stand. On Monday she used 4 bags of oranges. On Tuesday she used 1/2 as many oranges as on Monday.
Brut [27]
2 bags of oranges, 4/2=2
4 0
2 years ago
A die is rolled 20 times. given that three of the rolls came up 1, five came up 2, four came up 3, two came up 4, three came up
Serjik [45]

Solution:

Number of times a die is rolled = 20

1 - 3=A

2 - 5=B

3 - 4=C

4 - 2=D

5 - 3=E

6 -  3=F

Total number of arrangements of outcomes , when a dice is rolled 20 times given that 1 appear 3 times, 2 appears 5 times, 3 appear 4 times, 4 appear 2 times , 5 appear three times, and 6 appear 3 times

            = Arrangement of 6 numbers (A,B,C,D,E,F) in 6! ways and then arranging outcomes

= 6! × [ 3! × 5! × 4!×2!×3!×3!]

= 720 × 6×120×24×72→→[Keep in Mind →n!= n (n-1)(n-2)(n-3)........1]

= 895795200  Ways

8 0
3 years ago
Work out the value of: 5x - 2y when x = 3 and y = 4
KATRIN_1 [288]
The expression is 5x - 2y
Use PEMDAS and work out the rest
x = 3 and y = 4
5 x = 5 x 3 = 15
2 y = 2 x 4 = 8
15( x ) - 8 (y) = 7
The answer is 7

Hope This Helps You!
Good Luck Studying :)
4 0
3 years ago
Read 2 more answers
Math-- Farmer Bill has 500 meters of fencing and wants to enclose a rectangular plot that borders on a river. If farmer Bill doe
bixtya [17]

Answer:

Required dimensions of the rectangle are L = 200 m, W  = 100 m

The  largest area that can be enclosed is 20,000 sq m.

Step-by-step explanation:

The available length of the fencing = 500 m

Now, Perimeter of a rectangle = SUM OF ALL SIDES  = 2(L+B)

But, here once side of the rectangle is NOT FENCED.

So, the required perimeter  

= Perimeter of Complete field - Boundary of 1 open side

= 2(L+ W)   - L  = 2W + L

Now, fencing is given as 500 m

⇒  2W + L  = 500

Now, to maximize the length and width:

put L = 200, W = 100

we get 2(W) +L =  2(200) + 100 = 500 m

Hence, required dimensions of the rectangle are L = 200 m, W  = 100 m

The maximized area = Length x Width

                                   = 200 m x 100  m = 20, 000 sq m

Hence, the  largest area that can be enclosed is 20,000 sq m.

6 0
3 years ago
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