What is the answer to the fraction equation: 2/3+5/6÷1/3?

Hi can you please help me i cant answer this question.

Alex_Xolod [135]

Answer:

take a actual picture maby

3 0

3 years ago

Jermayne had 30 Apple he give 2/3 pf the apple to kath how many did he have left .

aleksklad [387]

She gave 2/3 so she has 1-2/3=1/3 left.

that makes 30*1/3=30/3=10 apples left

8 0

3 years ago

Read 2 more answers

Candice spent 5 1/4 hours doing her homework. Her brother, Ronald, spent 1/2 that number of hours doing his homework. How many h

Elden [556K]

Let

x-----> the number of hours that Candice spent doing her homework

y-----> the number of hours that Ronald spent doing his homework

we know that

$x=5\frac{1}{4}\ hours$

convert mixed number to an improper fraction

$5\frac{1}{4}\ hours=\frac{5*4+1}{4}= \frac{21}{4}\ hours$

so

$x=\frac{21}{4}\ hours$ -------> equation A

$y=\frac{1}{2}x$ ------> equation B

Substitute equation A in equation B

$y=\frac{1}{2}*\frac{21}{4}=\frac{21}{8}\ hours$

convert an improper fraction to a mixed number

$\frac{21}{8}\ hours=2.625=2+0.625=2+\frac{5}{8}=2\frac{5}{8}\ hours$

therefore

the answer is the option B

$2\frac{5}{8}\ hours$

5 0

3 years ago

Read 2 more answers

Determine the formula for the nth term of the sequence: -2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.