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2 years ago

What is the mathematical sentence of "A number increasd by 7 is 15?

Mathematics

1 answer:

svet-max [94.6K]2 years ago

6 0

Answer:

Step-by-step explanation:

A number increased by 7 means, 7 was added to the number and the result is 15

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The wholesale price for a shirt is $5.50. A certain department store marks up the wholesale price by 40%. Find the price of the

Kazeer [188]

Answer:

$7.7

Step-by-step explanation:

Hello, I can help you with this

you can solve this by using a simple rule of three.

Step 1

define

if the wholesale price for a shirt is $5.50,it is 100 percent

$5.50⇔100%

the new price(x) is 40% more than the original, it means 140%

x ⇔ 140%

the relation is

$\frac{5.5}{100}= \frac{x}{140}\\$

Step 2

solve for x(isolate x)

$\frac{5.5}{100}= \frac{x}{140}\\x=\frac{5.5*140}{100}\\x=7.7$

the price of the shirt in the department store is $7.7

Have a nice day.

7 0

3 years ago

The base of a rectangle is 15cm the perimeter is at most 94 cm. Write and solve an inequality to find the possible heights of th

tino4ka555 [31]

15 x 6 = 90 + 4 = 94 that's what I would do

4 0

3 years ago

a wrestler is weighing in for their match this evening. their normal weight is 168 pounds, but that weight can vary by as much a

Natalija [7]

Introduction

Wrestling is a wonderful activity with many advantages for the student-athlete. It is a sport that is highly

competitive, exciting and satisfying. It is a sport that provides for individual and team competition. It is -

and should be - fun. Unfortunately, the practice of losing weight by not eating, restricting fluid intake and

over-exercising reduces the sport's fun. This information is presented to help clear up misconceptions

regarding wrestling and weight loss. I also hope to give some guidance to those who desire to manage

their weight properly in preparation for and during the wrestling season.

History and Stigma

For too long, the wrestling community has unthinkingly accepted the myth that to be a good wrestler, you

must cut weight. The generally accepted thinking is something like this: if your natural weight is 135

pounds, you may be a good wrestler at 135 pounds. But if you wrestle at 130 pounds, you'll be a better

wrestler. And if you can make it down to 125, you'll be a state champion. No facts support that widely held

view, yet wrestlers and parents subscribe to that faulty reasoning. Looking further back, many remember

the days that losing excessive weight was a specific practice and expectation among wrestlers. It was

supposed to teach sacrifice, commitment, and the idea of “No Pain, No Gain.”

Regardless of the current attitude of the majority of the wrestling community, the stigma that an unhealthy

loss of weight is a requirement of wrestling among outside observers sticks. The Virginia High School

League has followed national guidelines to establish rules that encourage healthy weight management

among wrestlers.

Current Regulations

Preseason weight certification is accomplished with three steps. The first is determining each wrestler’s

body fat percentage using skin fold calipers. Next, the wrestler completes a hydration test, to insure

against a dehydrated weight measurement, and the wrestler weighs in. The third step is the calculation of

the wrestler’s minimum wrestling weight based on 7% body fat for males and 12% for females. The

wrestler may not wrestle at a weight class below his minimum weight during the season. The wrestler is

also restricted from losing more then 1.5% body weight loss per week (official weigh-ins at certification,

matches, and tournaments are used for this calculation). A one-pound per month growth allowance is

provided to allow for the natural growth of this age group. The VHSL advises, and we have instituted,

daily weigh-ins before and after practice to monitor weight-loss and dehydration.

Additionally, the National Federation of State High School Associations has adopted several other rules to

guide coaches and wrestlers while managing their weight. Wrestlers are discouraged from wrestling at a

weight class more then one weight class above their certified weight.

5 0

3 years ago

The symbol used to show that you are taking a root of something

konstantin123 [22]

the little check with the dash on top

8 0

3 years ago

Read 2 more answers

Find the taylor series centered at the given value of a and find the associated radius of convergence. (1) f(x) = 1 x , a = 1 (2

Kitty [74]

a) The radius of convergence is calculated as

R=1.

b) Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

<h3>What is the associated radius of convergence.?</h3>

(a)

Take into consideration the function f with respect to the number a,

$f(x)=\frac{1}{x}, \quad a=1$

In case you forgot, the Taylor series for the function $f$ at the number a looks like this:

$\begin{aligned}f(x) &=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n} \\&=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{*}(a)}{2 !}(x-a)^{2}+\ldots\end{aligned}$

Determine the function f as well as any derivatives of the function $f by setting a=1 and working backward from there.

$\begin{aligned}f(x) &=\frac{1}{x} & f(1)=\frac{1}{1}=1 \\\\f^{\prime}(x) &=-\frac{1}{x^{2}} & f^{\prime}(1)=-\frac{1}{(1)^{2}}=-1 \\\\f^{\prime \prime}(x) &=\frac{2}{x^{3}} & f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2 \\\\f^{\prime \prime}(x) &=-\frac{2 \cdot 3}{x^{4}} & f^{\prime \prime}(1)=-\frac{2 \cdot 3}{(1)^{4}}=-2 \cdot 3 \\\\f^{(*)}(x) &=\frac{2 \cdot 3 \cdot 4}{x^{5}} & f^{(n)}(1)=\frac{2 \cdot 3 \cdot 4}{(1)^{5}}=2 \cdot 3 \cdot 4\end{aligned}$

At the point when a = 1, the Taylor series for the function f looks like this:

$f(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+\cdots \\\\&=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime}(1)}{3 !}(x-1)^{3}+\cdots \\$

$&=1+\frac{-1}{1 !}(x-1)+\frac{2}{2 !}(x-1)^{2}+\frac{-2 \cdot 3}{3 !}(x-1)^{3}+\frac{2 \cdot 3 \cdot 4}{4 !}(x-1)^{4}+\cdots \\\\&=1-(x-1)+(x-1)^{2}-(x-1)^{3}+(x-1)^{4}+\cdots \\\\&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

In conclusion,

$&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

Find the radius of convergence by using the Ratio Test in the following manner:

$\begin{aligned}L &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \\&=\lim _{n \rightarrow \infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{(-1)^{n}(x-1)^{n}} \mid \\&=\lim _{n \rightarrow \infty}|x-1| \\&=|x-1|\end{aligned}$

The convergence of the series when L<1, that is, |x-1|<1.

The radius of convergence is calculated as

R=1.

For B

Take into consideration the function f with respect to the number a,

$a_{n}=(-1)^{n}(x-1)^{n}$

$f(x)=\left(x^{2}+2 x\right) e^{x}, a=0$ The Taylor series for f(x)=e^{x} at a=0 is,

$e^{2}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots$

$f(x) &=\left(x^{2}+2 x\right) e^{x} \\&=\left(x^{2}+2 x\right)\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+2 x\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right) \\&=x^{2}\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+\left(\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\frac{x^{6}}{4 !}+\ldots\right)+\left(2 x+2 x^{2}+\frac{2 x^{3}}{2 !}+\frac{2 x^{4}}{3 !}+\frac{2 x^{5}}{4 !}+\ldots\right) \\$

$&=\left(x^{2}+x^{3}+\frac{x^{4}}{2 !}\right) \\&=2 x+3 x^{2}+\left(1+\frac{2}{2 !}\right) x^{3}+\left(\frac{1}{2 !}+\frac{2}{3 !}\right) x^{4}+\left(\frac{1}{4 !}\right) x^{5}+\ldots \\&=2 x+3 x^{2}+2 x^{3}+\frac{5}{6} x^{4}+\frac{1}{4} x^{5}+\ldots$

Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.