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1 year ago

At a restaurant a meal used to cost $9, but a new owner decided to raise the price by 13

Mathematics

1 answer:

kari74 [83]1 year ago

6 0

Answer:

$10.17

Step-by-step explanation:

9 x .13 = 1.17

9 + 1.17 = 10.17

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There are 234 students in 9 different classrooms. What is the ratio of students to classrooms?

Lina20 [59]

Answer:

26 students in each classroom

Step-by-step explanation:

234/9=26

6 0

3 years ago

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The formula p=f/a can be used to relate pressure, force and area. solve the formula for a

Gemiola [76]

<h3>Answer: a = f/p</h3>

===============================================

Explanation:

We'll solve for 'a' doing these two steps in this exact order

Multiply both sides by 'a'
Divide both sides by p

We do this because we're following PEMDAS in reverse to isolate the variable we want.

The steps look like this

p = f/a

pa = f

a = f/p

Note how ultimately the 'a' and p variables swap places when going from p = f/a to a = f/p

3 0

3 years ago

Simplify.<br><br> 4−16÷4+42<br><br> −16<br> 13<br> 15<br> 16<br>PLEASE HELP

jok3333 [9.3K]

PEMDAS. 16/4 = 4. 4+42 = 46. 46-4 = 42. Answer is 42!

3 0

3 years ago

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Now, lets evaluate the same integral using power series. first, find the power series for the function f(x = \frac{32}{x^2+4}. t

BlackZzzverrR [31]

No idea what the previous part of the problem is, but you have

$f(x)=\dfrac{32}{x^2+4}=\dfrac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_{n\ge0}\left(-\frac{x^2}4\right)^n$
$f(x)=\displaystyle8\sum_{n\ge0}\left(-\dfrac14\right)^nx^{2n}$

which is valid for $\left|-\dfrac{x^2}4\right|$ , or $|x|$ . So the integral from 0 to 2 is

$\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_{n\ge0}\left(-\frac14\right)^nx^{2n}\,\mathrm dx$
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\int_0^2x^{2n}\,\mathrm dx$

Note that since the power series only converges on the interval if $x$ is strictly less than 2, which means we have to treat this as an improper integral.

$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\int_0^tx^{2n}\,\mathrm dx$ [/tex]
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}$
$=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}$
$=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}$
$=16-\dfrac{16}3+\dfrac{16}5-\dfrac{16}7+\dfrac{16}9+\cdots$

6 0

3 years ago

What is the answer to -4.3g=25.8

8090 [49]

G=-6
You just divide both sides by -4.3

7 0

3 years ago