Can you guys help with this one please, thanks

Mathematics

1 answer:

densk [106]3 years ago

7 0

5^6 (5 to the 6th power)

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A store sells mugs. Each mug costs the same amount. During a sale, the store reduces the price of each mug by $1.75. Wen spends

Sergio039 [100]

I’m assuming the question is what is the price of one mug WITHOUT the sale.

So since the price of 5 mugs is $28.20, we can work out the price of one mug with the sale.

We do this by doing:

$28.20 / 5

Which gives us:

$5.64 per mug during the sale.

To work out the price of a mug without the sale, we simply add on the amount that has been taken off (which we know is $1.75 because the question says so):

$5.64 + $1.75 = $7.39

Therefore we can conclude the price of a mug without a sale is $7.39

And the price of a mug during the sale is $5.64.

I hope this helps!

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There is 1/3 gallon of water in a 3-gallon container. What fraction of the container is filled?

mina [271]

Answer:

ONE-THIRD OF A GALLON DIVIDED BY 3 GALLONS EQUALS ONE-NINTH

Step-by-step explanation:=1/9

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How can patterns help me divide multiples of 10

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When you lay them out then you know what you are dividing

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Identify the corresponding word problem given the equation: x + 12.50 = 55.75 A) For mowing the neighbor's yard, Jessy was paid

sergejj [24]

The answer is A. because for you to find the answer you have to subtract so it would be 55.75-12.5=43.25

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Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum

lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

$\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\$

$\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\ (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\$

$=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3} \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\$

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2 years ago