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Oksi-84 [34.3K]
2 years ago
11

The cost of an item is 100% of its price. What number goes in place of ? in the addition problem?

Mathematics
2 answers:
Dmitry_Shevchenko [17]2 years ago
8 0
I believe this is the complete problem. 

<span>Anton bought a picnic cooler. His total bill, with tax, was $7.95. He paid 6 percent sales tax. How much did he pay for the cooler alone without the tax? The cost of an item is 100% of its price. What number goes in place of ? in the addition problem?
</span>
The answer is the below:

<span>He paid $7.473 without tax because $7.95 x 96% or (0.96) =$7.473</span>
borishaifa [10]2 years ago
6 0

Answer:

The value of x is $7.5

Step-by-step explanation:

Given The cost of an item is 100% of its price.

Anton bought a picnic cooler. The total bill, with the tax, was $7.95. He paid 6% sales tax.

we have to tell how much he pay for cooler without tax.

We have to find the value of ? let it be x shown in expression.

By the given expression we can write

x+6% of x=$7.95

⇒ x(1+0.06)=$7.95

⇒ 1.06x=7.95 ⇒ x=7.5

Hence, the value of x is $7.5

He paid $7.5 without taxes.

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Circle F has a diameter of 36m and = 240o. Find the length of . Leave your answer in terms of pi.
andrezito [222]

Answer:

So, the arc length is 24\pi

Step-by-step explanation:

We are given

diameter =36m

so, we can find radius

r=\frac{36}{2}=18m

and angle as

\theta=240

Since, it is in degree

so, we can change it in radian

\theta=240\times \frac{\pi}{180}

\theta=\frac{4\pi}{3}

now, we can find arc length

L=r\times \theta

now, we can plug values

L=18\times \frac{4\pi}{3}

L=24\pi


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\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.

\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

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