1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
katen-ka-za [31]
3 years ago
5

1 It takes 1 cups of sugar and 3 cups of flour

Mathematics
1 answer:
scZoUnD [109]3 years ago
8 0

For every 1 cup of sugar there's 3 cups of flour or vice versa. (For every 3 cups of flour there's 1 cup of sugar.)

<h2>Answer:</h2>

<u>The ratio of sugar to flour is </u><u>1 to 3</u><u>, </u><u>1:3</u><u>, or </u><u>1/3</u><u>.</u>  (Option B)

Hope this helps :)

You might be interested in
It takes Carl 2/3 of an hour to clean his room every day.How many hours did he spend cleaning his room in the month of May?
MrMuchimi
20 2/3 hrs or 20 hours and 40 minutes cleaning his room. There are 30 months in may, so you multiply this by 2/3 to get 20 2/3 hrs
7 0
3 years ago
Read 2 more answers
Factor the equation 2x^2+5x+3
lozanna [386]

Answer:

(2x+3)(x+1)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Use prime factorization to reduce the fraction 36/180
laila [671]

Answer:

Step-by-step explanation:

36 = 6* 6 = 2* 3 * 2* 3 = 2*2*3*3

180 = 2 * 2 * 3 * 3 * 5 = 2 * 2 * 3*3*5

\frac{36}{180}= \frac{2*2*3*3}{2*2*3*3*5} =  \frac{1}{5}

Factors that are common will be cancelled

5 0
2 years ago
Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]​
Schach [20]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

\rm :\longmapsto\:n +  {n}^{2} +  {n}^{3}  +  -  -  -  +  {n}^{n}

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

\rm \:  =  \: \dfrac{n( {n}^{n}  - 1)}{n - 1}

\rm \:  =  \: \dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }

Now, Consider Denominator, we have

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {n}^{n}

can be rewritten as

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {(n - 1)}^{n} +   {n}^{n}

\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

Now, Consider

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

So, on substituting the values evaluated above, we get

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{ {n}^{n}  - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

Now, we know that,

\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x}  =  {e}^{k}}}}

So, using this, we get

\rm \:  =  \: \dfrac{1}{1 +  {e}^{ - 1}  + {e}^{ - 2} +  -  -  -  -  \infty }

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

\rm \:  =  \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }

\rm \:  =  \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }

\rm \:  =  \: \dfrac{1}{\dfrac{e}{e - 1} }

\rm \:  =  \: \dfrac{e - 1}{e}

\rm \:  =  \: 1 - \dfrac{1}{e}

Hence,

\boxed{\tt{ \displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} } =  \frac{e - 1}{e} = 1 -  \frac{1}{e}}}

3 0
2 years ago
Find f^-1(x) for f(x) = 1/x^3
kow [346]

Answer: f^{-1}(x) = \frac{\sqrt[3]{x^{2}}}{x}

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

y = \frac{1}{x^{3}}

Inverse is when you swap the x's and y's and then solve for "y":

x = \frac{1}{y^{3}}

y^{3} = \frac{1}{x}

y = \frac{1}{\sqrt[3]{x}}

y = \frac{1}{\sqrt[3]{x}}*(\frac{\sqrt[3]{x}}{\sqrt[3]{x}})^{2}

y = \frac{\sqrt[3]{x^{2}}}{x}

4 0
3 years ago
Other questions:
  • Which statement is true about the difference 2√7-√28
    14·1 answer
  • I need help with Tuesday hw
    10·1 answer
  • Write 27/64 in the simplest form
    8·2 answers
  • Write a rule to describe each transformation
    9·1 answer
  • Do 1/3 and 2/6 name the same part of one whole?
    12·2 answers
  • A hawk soars at an altitude of 1,800ft. If the hawk descends to the ground in 45 min, what is the vertical speed?
    5·2 answers
  • Sofia invests her money in an account paying 7% interest compounded semiannually. What is the effective annual yield on this acc
    14·1 answer
  • (3, -3); m = -2<br> In standard form
    12·1 answer
  • 2.5x + 8 = 1.5x − 12<br> What is the value of x?
    15·1 answer
  • Factorise completely 12 x³y - 18 xy²​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!