Answer: 431 reserved seats were sold.
Step-by-step explanation:
<h3>4(1787-y)+3y = 5792
</h3><h3>7148-4y+3y = 5792</h3><h3>x+1356=1787
</h3><h3>x=431</h3><h3 /><h3>Hope this helps :)</h3>
Jeff should purchase 25 , 2 ounce box and it will cost him $27.5 .
<h3>What is an ounce ?</h3>
Ounce is defined as 1/16 part of a pound.
It is given that
price of a 2-ounce box is $1.10,
a 16-ounce box is $8.98
a 32-ounce box is $17.98.
As Jeff wants to purchase total 48 ounce chocolates , the box which is more cheaper has to be determined
The price of 1 ounce in a 2 ounce box is $1.1 /2 = $ 0.55
The price of 1 ounce in a 16 ounce box is $8.98 /16 = $ 0.56
The price of 1 ounce in a 32 ounce box is $17.98 /32 = $ 0.56
Therefore Jeff should purchase 25 , 2 ounce box and it will cost him $27.5 .
To know more about Ounce
brainly.com/question/26950819
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When Javier writes 40 on both accounts they cost him the same amount of money so it should be 41.
Javier needs to write 41 checks for the second bank to beat better
Answer:
![\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B1.%5C%20f%5E%7B-1%7D%28x%29%3D4%5Clog%28x%5Csqrt%5B4%5D2%29%7D%5C%5C%5C%5C%5Cboxed%7B2.%5C%20f%5E%7B-1%7D%28x%29%3D%5Clog%28x%5E5%2B5%29%7D%5C%5C%5C%5C%5Cboxed%7B3.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B4%5E%7Bx-1%7D%7D%7D)
Step-by-step explanation:
![\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)](https://tex.z-dn.net/?f=%5Clog_55%5E%7B%5Cfrac%7B1%7D%7B4%7Dy%7D%3D%5Clog_5%5Cleft%282%5E%5Cfrac%7B1%7D%7B4%7Dx%5Cright%29%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%5Cfrac%7B1%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B4%7Dy%3D%5Clog%28x%5Csqrt%5B4%5D2%29%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%204%7D%5C%5C%5C%5Cy%3D4%5Clog%28x%5Csqrt%5B4%5D2%29)
![--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)](https://tex.z-dn.net/?f=--------------------------%5C%5C2.%5C%5Cy%3D%2810%5Ex-5%29%5E%5Cfrac%7B1%7D%7B5%7D%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C%2810%5Ey-5%29%5E%5Cfrac%7B1%7D%7B5%7D%3Dx%5Cqquad%5Ctext%7B5%20power%20of%20both%20sides%7D%5C%5C%5C%5C%5Cbigg%5B%2810%5Ey-5%29%5E%5Cfrac%7B1%7D%7B5%7D%5Cbigg%5D%5E5%3Dx%5E5%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%2810%5Ey-5%29%5E%7B%5Cfrac%7B1%7D%7B5%7D%5Ccdot5%7D%3Dx%5E5%5C%5C%5C%5C10%5Ey-5%3Dx%5E5%5Cqquad%5Ctext%7Badd%205%20to%20both%20sides%7D%5C%5C%5C%5C10%5Ey%3Dx%5E5%2B5%5Cqquad%5Clog%5C%20%5Ctext%7Bof%20both%20sides%7D%5C%5C%5C%5C%5Clog10%5Ey%3D%5Clog%28x%5E5%2B5%29%5CRightarrow%20y%3D%5Clog%28x%5E5%2B5%29)
Fraction does not make sense this question
