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2 years ago

50 is 25% of what number

Mathematics

2 answers:

Tatiana [17]2 years ago

8 0

look it up on the internet

It is right there

Marat540 [252]2 years ago

5 0

200. 50x4=200 hope that helps

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Can someone help wit this asap i have 10 more mins

bazaltina [42]

A is the correct answer

4 0

3 years ago

Karen gaines invested $15,000 in a money market account with an interest rate of 2.25% compounded semiannually. Five years later

fredd [130]

Answer:

years later , karen withdrew the full amount to put toward the down payment on a new house. How much did she withdraw? ... gaines invested $15,000 in a money market account with an interest rate of 2.75% compounded semiannually. ... Median response time is 34 minutes and may be longer for new subjects.

6 0

3 years ago

Solve the equation 29 - (x+8) = 6x - 7

max2010maxim [7]

First step you should do is simplify both sides of your equation:
29-(x+8)=6x-7
Distribute the Negative Sign:
29+-1(x+8)=6x-7
29+-1x+(-1)(8)=6x-7
29+-x+-8=6x-7
29+-x+-8=6x+-7
Combine Like Terms:
(-x)+(29+-8)=6x-7
-x+21=6x-7
Subtract 6x from both sides:
-x+21-6x=6x-7-6x
-7x+21=-7
Subtract 21 from both sides:
-7x+21-21=-7-21
-7x=-28
Divide both sides by -7:
-7x/-7 = -28/-7
And now your answer should be:
x=4
~~~~~
Good luck~ Sans

6 0

2 years ago

Brown has own bakery he baked 5 cakes per day due to occasional christmas story to be in the whole christmas week how many cakes

Lena [83]

she would make 35 cakes

8 0

3 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

3 years ago