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insens350 [35]
3 years ago
15

Bottles of sparkling water usually cost $1.69. The grocer changed the price to 4 for $5. You bought one bottle last week for $1.

69 and one bottle this week at the 4 for $5 price. Did you pay more or less for the bottle this week at 4 for $5? How much more or less?
Mathematics
2 answers:
sleet_krkn [62]3 years ago
8 0

Given :

Cost of single bottle, C = $1.69 .

Cost of 4 bottles, B = $5 .

You bought one bottle last week for $1.69 and one bottle this week at the 4 for $5 price.

To Find :

Did you pay more or less for the bottle this week at 4 for $5? How much more or less?

Solution :

Effective price per bottle if we purchase 4 is, E = 5/4 = $1.25 .

Regular price, R = $1.69 .

Change in price, C = $( 1.69 - 1.25 ) = $0.44 .

Therefore, we are paying $0.44 less than the actual price.

Hence, this is the required solution.

babymother [125]3 years ago
6 0

Answer:0.44

Step-by-step explanation:

you divide 5 by 4 and then multiply 1.25 by 4 after subtract 1.69 by 0.44

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Answer:

33/6

Step-by-step explanation:

Convert the proper fraction into an improper fraction.

11/2

Multiply numerator and denominator by the same number. This number needs to allow the demominator to be 6.

(11*3)/(2*3)

33/6

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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
Can you help me<br>4(b–1)²–9​
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Answer:

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= 4(b² - 2b + 1) - 9

= 4b² - 8b + 4 - 9

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Hope this helps!

7 0
2 years ago
please explain your answer( btw me and my family ahre an account so if u see omebody answer an question without the answer its m
lozanna [386]
I think the answer is 15,840!
3 0
3 years ago
Read 2 more answers
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