All categories

3 years ago

1. What percent of 56 is 14?

Mathematics

2 answers:

maksim [4K]3 years ago

5 0

<u>Answer:</u>

<u>Step-by-step explanation:</u>

14 ÷ 56 × 100

tekilochka [14]3 years ago

3 0

Answer:

Percentage Calculator: 14 is what percent of 56? = 25.

Step-by-step explanation:

You might be interested in

<img src="https://tex.z-dn.net/?f=x%20%5C6%20-%201%20%3D%20%20x%20%5C4%20%2B%201%20%5C2" id="TexFormula1" title="x \6 - 1 = x \

skelet666 [1.2K]

Answer:

-18

Step-by-step explanation:

Hope it helps....................................................

6 0

2 years ago

A glass is 1/4 full. Then 24cm of orange juice is poured in. The glass is now 2/5 full. What is the total volume of the glass?

Phantasy [73]

160cm3

2/5 - 1/4 = 8/20 - 5/20 = 3/20

(3/20)t = 24
t = 160

6 0

2 years ago

An ice chest contains six cans of apple juice, eight cans of grape juice, four cans of orange juice, and two cans of mango ju

Katena32 [7]

6+8+4+2=20
2/20 to grab mango can = 1/10
8/19 (remove one for previous) fro grape juice
6/18 for apple juice = 1/3
1/10*8/19*1/3
=
8/570
=
4/285

8 0

3 years ago

PLEASE HELP ME!! LAST ASSIGNMENT

Darya [45]

Answer:

u = 3

Step-by-step explanation:

$\frac{10}{14} = \frac{35}{u + 46} \\\\ 10(u + 46) = 35×14 \\\ 10u + 460 = 490 \\\ 10u = 490 - 460 \\\ 10u = 30 \\\ u = \frac{30}{10} \\\\ u = 3$

I hope I've helped you.

5 0

2 years ago

Read 2 more answers

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago