)) Matt's bill for breakfast at a restaurant was $91. He left an 18% tip. What was the total

Mathematics

2 answers:

gladu [14]3 years ago

5 0

107.38 you take 18 percent of 91 then add it to 91 getting your final total

gladu [14]3 years ago

4 0

Answer:

$107.38

Step-by-step explanation:

$91 (the price) multiplied by the percentage 18% (91 x 18 = 1630) (1638/100=16.38 [the tip})

$91 (th original price) + 15.38 (the tip) = $108.38 (final price)

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Can someone solve this? and tell me how?

maksim [4K]

Answer:

1/3

Step-by-step explanation:

When working with balanced expressions (stuff on both sides of the equal sign), "what you do to one side, you do to the other", which keeps it balanced.

The first thing we notice is the exponent 1/4, which is one both sides, so we can get rid of it on both sides by using the reverse operation.

The reverse of exponents is square root.

$(4x + 10)^{\frac{1}{4}} = (9 + 7x)^{\frac{1}{4}}\\\sqrt[\frac{1}{4}]{(4x + 10)^{\frac{1}{4}}} = \sqrt[\frac{1}{4}]{(9 + 7x)^{\frac{1}{4}}}\\\\4x + 10 = 9 + 7x$

Isolate x to solve. Separate the variables and non-variables.

4x + 10 = 9 + 7x

4x - 4x + 10 = 9 + 7x - 4x Subtract 4x from both sides

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10 - 9 = 9 - 9 + 3x Subtract 9 from both sides

1 = 3x Divide both sides by 3 to isolate x

x = 1/3 Answer

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2 years ago

The value of x is -1/4. Order the expressions from least to greatest. Enter 1 for the least amount; 2 for the second least amoun

Natasha_Volkova [10]

Answer:

1/20 trust me

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the half range Fourier sine series of the function

worty [1.4K]

The full range is $-\pi$ (length $2L=2\pi$ ), so the half range is $L=\pi$ . The half range sine series would then be given by

$f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx$

where

$b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx$

Essentially, this is the same as finding the Fourier series for the function

$\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0$

Integrating by parts yields

$b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n$

So the half range sine series for this function is simply

$f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n$