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Stolb23 [73]
3 years ago
12

What is the intermediate step in the form (x+a)2 = b as a result of completing the square for the following x^2+16x-145=10x-7

Mathematics
1 answer:
Lorico [155]3 years ago
7 0

Answer:

(x+3)^2=147

Step-by-step explanation:

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Answer:

y=-3x-3

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Please help with this question.
katrin2010 [14]
The answer is 10 minutes
7 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
3 years ago
Jamie has a collection of 78 nickels, dimes, and quarters worth $12.40. If the number of quarters is
fgiga [73]

Jamie has 25 nickels, 14 dimes and 39 quarters in his collection

Step-by-step explanation:

The given is:

1. Jamie has a collection of 78 nickels, dimes, and quarters

2. They worth $12.40

3. If the number of quarters is  doubled, the value becomes $22.15

Assume that there are n nickels, d dimes and q quarters

∵ The collection of Jamie has 78 coins

∴ n + d + q = 78 ⇒ (1)

∵ 1 nickels = 5 cents

∵ 1 dimes = 10 cents

∵ 1 quarter = 25 cents

∵ The collection of Jamie worth $12.40

- Change the value of the collection to cents

∵ $1 = 100 cents

∴ The collection of Jamie worth = 12.40 × 100 = 1240 cents

∴ 5n + 10d + 25q = 1240

- Divide all terms by 5 to simplify it

∴ n + 2d + 5q = 248 ⇒ (2)

∵ When the number of quarters is  doubled, the value becomes

   $22.15

∵ The number of quarters is q

∴ The new number of quarters is 2q

∵ The value of coins is $22.15

∵ $22.15 = 22.15 × 100 = 2215 cents

∴ 5n + 10d + 25(2q) = 2215

∴ 5n + 10d + 50q = 2215

- Divide each term by 5 to simplify it

∴ n + 2 d + 10q = 443 ⇒ (3)

Subtract equation (2) from equation (3)

∵ n + 2d + 5q = 248 ⇒ (2)

∵ n + 2 d + 10q = 443 ⇒ (3)

∴ 5q = 195

- Divide both sides by 5

∴ q = 39

Substitute the value of q in equations (1) and (2)

∵ n + d + q = 78 ⇒ (1)

∴ n + d + 39 = 78

- Subtract 39 from both sides

∴ n + d = 39 ⇒ (4)

∵ n + 2d + 5q = 248 ⇒ (2)

∴ n + 2d + 5(39) = 248

∴ n + 2d + 195 = 248

- Subtract 195 from both sides

∴ n + 2d = 53 ⇒ (5)

Subtract equation (4) from equation (5)

∵ n + d = 39 ⇒ (4)

∵ n + 2d = 53 ⇒ (5)

∴ d = 14

Substitute the value of d in equation (4)

∵ n + d = 39 ⇒ (4)

∴ n + 14 = 39

- Subtract 14 from both sides

∴ n = 25

Jamie has 25 nickels, 14 dimes and 39 quarters in his collection

Learn more:

You can learn more about solving system of equations in

brainly.com/question/13168205

brainly.com/question/2115716

#LearnwithBrainly

5 0
3 years ago
Select the expression that is equivalent to ( x + 4 )^2
dangina [55]

The equivalent expression of (x+4)² is x²+4x+16.

<h3>What is an expression?</h3>

Expression in mathematics is combination of variables with the use of operations and given rules. It can be in the form of equation, numbers etc.

The given expression is (x+4)².

Simplifying the expression;

(x+4)²

= (x+4) (x+4)

= x²+4x+16

Therefore, the equivalent expression of (x+4)² is x²+4x+16.

To learn more about the expression;

brainly.com/question/24242989

#SPJ1

3 0
1 year ago
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