All categories

3 years ago

What is the quotient of 6/7 and 3/14

Mathematics

2 answers:

Goryan [66]3 years ago

5 0

Answer: 4

Step-by-step explanation: To divide fractions, you have to switch the operation to multiplication, and multiply the first fraction by the reciprocal of the second fraction. So 6/7 times 14/3. Cross cancel out, so 14 and 7. 7 cancels out to be 1, 14 cancels out to be two. 6 and 3, are the same thing. So in the end you have 2/1 times 2/1 which is 4.

lbvjy [14]3 years ago

3 0

Answer:

6/7 divided by 3/14=4

Step-by-step explanation:

Hope this helps!

You might be interested in

1.1x+1.2x−5.4=−10 x=

katrin [286]

1.1x+1.2x-5.4=-10x

Move all terms to the left:

1.1x+1.2x-5.4-(-10x)=0

Add all the numbers together, and all the variables

2.3x-(-10x)-5.4=0

Get rid of parentheses

2.3x+10x-5.4=0

Add all the numbers together, and all the variables

12.3x-5.4=0

Move all terms containing x to the left, all other terms to the right

12.3x=5.4

x=5.4/12.3

x=3.8571428571429/8.7857142857143

Hope it helps!

6 0

3 years ago

I need some help with 22, 24, 25

valkas [14]

22 increases by 0.9
24 increases by 4 and 1/2
25 increases by 4.3
Just subtract the number from the number after it to find how to find the next numbers in the sequence

5 0

3 years ago

Read 2 more answers

What is the minimum value of the function g (2) = x2 - 6x – 12?<br> -21<br> 3+ V21<br> 3<br> 3-21

kirill [66]

Answer:

g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

4x2 - 5 + x + 5x2 + 3x - 9

Mnenie [13.5K]

Answer:

9x2+4x−14

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:

BabaBlast [244]

Answer

$a=0$ , $b=2$

$g_1(x)=\frac{5x}{2}$ , $g_2(x)=7-x$

Step-by-step explanation:

Given that

$\int \int Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)$

For the term $\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=\frac {2y}{5}$ and for $y$ is from $y=0$ to $y=5$ and the region $D$ , for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=\frac{5x}{2}$ to $y=5$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

$\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)$

Similarly, for the other term $\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=7-y$ and limits for $y$ is from $y=5$ to $y=7$ and the region $D$ , for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=5$ to $y=7-x$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

$\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)$

Hence, from equations $(i)$ , $(ii)$ and $(iii)$ , on reversing the order of integration, the required expression is

$\int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)$

Now, compare the RHS of the equation $(iv)$ with

$\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx$

We have,

$a=0, b=2, g_1(x)=\frac{5x}{2}$ and $g_2(x)=7-x$ .

3 0

3 years ago