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KengaRu [80]
3 years ago
9

(2) Solve for x please help

Mathematics
1 answer:
goblinko [34]3 years ago
4 0
The answer is x is greater than 2
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If Nicholas is located at (–3, 2), what is the distance to get to the grocery store?
PilotLPTM [1.2K]

Answer:

it would be D 5 hope this helps :)

Step-by-step explanation:

5 0
3 years ago
Check-Ins 2 NC Math 1
fomenos

The maximum amount of profit the carnival makes based on the number of tickets sold is 6.5 thousand of dollars,

<h3>How to determine the difference?</h3>

The function is given as:

f(x) = -0.5x^2 + 5x - 6

Differentiate the function

f'(x) = -x + 5

Set to 0

-x + 5 = 0

Make x the subject

x = 5

Substitute x = 5 in f(x)

f(5) = -0.5 * 5^2 + 5 * 5 - 6

f(5) = 6.5

The table is not given.

Hence, the maximum amount of profit the carnival makes based on the number of tickets sold is 6.5 thousand of dollars,

Read more about quadratic functions at:

brainly.com/question/18797214

5 0
1 year ago
Please help me with the following questions. Thanks!
Natali5045456 [20]

Answer:

The answer to your questions  are in bold

Step-by-step explanation:

a)  

 C = \left[\begin{array}{ccc}-6&6\\-2&4\\\end{array}\right]

    = -24 + 12

   = -12

b)  -1   7          -4   -1                   -1  - 4        7 - 1               - 5      6

    -2 -6          -8    8      =          -2 - 8       -6 + 8              -10      2

     2 -3           2   -7                  2 + 2        -3 - 7               4       -10

   -1   10         -6     5                 -1 - 6         10 + 5              -7       15

5 0
3 years ago
10. Three kinds of teas are worth $4.60 per pound, $5.75 per pound, and $6.50 per pound. They are to be
zepelin [54]

Answer:

The mass of the $4.60/lb tea that should be used in the mixture is 10 lb

The mass of the $5.75/lb tea that should be used in the mixture is 8 lb

The mass of the $6.50/lb tea that should be used in the mixture is 2 lb

Step-by-step explanation:

The parameters of the question are;

The worth of the three teas are

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of the mixture of the three teas = 20 lb

The worth of the mixture of the three teas = $5.25 per pound = $5.25/lb

The amount of the $4.60 in the mixture = The sum of the amount of the other two teas

Therefore, given that the mass of the mixture = 20 lb, we have in the mixture;

The mass of tea A + The mass of Tea B + The mass of Tea C = 20 lb

The mass of tea A = The mass of Tea B + The mass of Tea C

Therefore;

The mass of tea A + The mass of tea A = 20 lb

2 × The mass of tea A in the mixture = 20 lb

The mass of tea A in the mixture = 20 lb/2 = 10 lb

The mass of tea A in the mixture = 10 lb

The mass of Tea B + The mass of Tea C = The mass of tea A = 10 lb

The mass of Tea B + The mass of Tea C = 10 lb

The mass of Tea B  = 10 lb - The mass of Tea C

Where the mass of Tea C in the mixture = x, we have;

The mass of Tea B in the mixture = 10 lb - x

The cost of the 10 lb of tea A = 10 × $4.60 = $46.0

The worth of the tea mixture = 20 × $5.25 = $105

The worth of the remaining 10 lb of the mixture comprising of tea A and tea B is given as follows;

The worth of Tea B + The worth of Tea C in the mixture = $105.00 - $46.00 = $59.00

Therefore, we have;

x lb × $6.50/lb + (10 - x) lb × $5.75/lb = $59.00

x × $6.50 - x × $5.75 + $57.50 = $59.00

x × $0.75 = $59.00 -  $57.50 = $1.50

x =  $1.50/$0.75 = 2 lb

∴ The mass of Tea C in the mixture = 2 lb

The mass of Tea B in the mixture = 10 lb - x = 10 lb - 2 lb = 8 lb

The mass of Tea B in the mixture = 8 lb

Therefore, since we have;

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of tea A in the mixture = 10 lb

The mass of tea B in the mixture = 8 lb

The mass of tea C in the mixture = 2 lb, we find;

The mass of the $4.60/lb tea that should be used in the mixture = 10 lb

The mass of the $5.75/lb tea that should be used in the mixture = 8 lb

The mass of the $6.50/lb tea that should be used in the mixture = 2 lb.

6 0
2 years ago
(x + 1)^2 = 3x – 1 solving Quadratics ?
Deffense [45]

Step-by-step explanation:

(x + 1)^{2}  = 3x - 1 \\  \\  \therefore \:  {x }^{2}  + 2x + 1 = 3x - 1 \\  \\ \therefore \:  {x }^{2}  + 2x + 1  -  3x  + 1 = 0\\  \\ \therefore \:  {x }^{2}  + 2x  -  3x + 1 - 1 = 0 \\  \\ \therefore \:  {x }^{2}  - x  = 0 \\  \\ \therefore \:  x(x - 1) = 0 \\  \\ \therefore \: x = 0 \:  \: or \:  \: x - 1 = 0 \\  \\ \therefore \: x = 0 \:  \: or \:  \: x = 1 \\  \\  \huge{ \red{ \boxed{\therefore \: x =  \{0, \:  \: 1 \}}}}

5 0
3 years ago
Read 2 more answers
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