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3 years ago

A restaurant bill will be paid equally between 4 friends. The bill totaled $25.20. How much should each friend pay?

Mathematics

2 answers:

ruslelena [56]3 years ago

8 0

Answer:

$6.30

Step-by-step explanation:

25.20 divided by 4.

Mama L [17]3 years ago

3 0

$6.30
25.20(dollars)/4(friends) = 6.30

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3 years ago

Why would someone want to factor a polynomial? Provide real-world examples of different questions we can answer or factors we ca

IgorC [24]

Factoring is a common mathematical process used to break down the factors, or numbers, that multiply together to form another number. Some numbers have multiple factors.

<u>Explanation:</u>

Factoring polynomials involves breaking up a polynomial into simpler terms (the factors) such that when the terms are multiplied together they equal the original polynomial. Factoring helps solve complex equations so they are easier to work with. Factoring polynomials includes: Finding the greatest common factor.

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3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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2 years ago

Adult tickets for the school play are $12.00 and children's tickets are $8.00. If a theatre holds 300 seats and the sold out per

Vaselesa [24]

Let x = number of adult tickets, and y = number of children tickets. One equation must deal with the number of tickets, and the other equation must deal with the revenue from the tickets.

Then x + y = 300 is the number of tickets
12x + 8y = 3280 is the revenue from the tickets.

Using the substitution method:

x + y = 300   ⇒   y = 300 - x   ⇒   Equation (3)

12x + 8y = 3280   ⇒   12x + 8(300-x) = 3280   ⇒   x = 220

y = 300 - x   ⇒   y = 300-220 ⇒ 80

Therefore 220 adult tickets and 80 children's tickets were sold.

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3 years ago

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dusya [7]

It would be the first graph because it lands on -1 and +4

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3 years ago

Read 2 more answers