All categories

2 years ago

3 3/8 divided by 3/4 using cross multiplication algorithm

Mathematics

1 answer:

Bogdan [553]2 years ago

5 0

Answer

3 3/8 ÷ (3/4) = (9/2) = 4 ½

Explanation

We are asked to evaluate 3 3/8 divided by 3/4 using cross multiplication algorithm.

The algorithm will need us to put the mixed fraction in definite fraction form (improper fraction) and to evaluate the division, we change the division sign into multiplication sign and the fraction after the division sign changes into its reciprocal.

3 3/8 = (27/8)

(3 3/8) ÷ (3/4)

= (27/8) ÷ (3/4)

$\begin{gathered} \frac{27}{8}\div\frac{3}{4} \\ =\frac{27}{8}\times\frac{4}{3} \\ =\frac{108}{24} \\ \text{Divide numerator and denominator by 12} \\ \frac{108}{24}=\frac{9}{2} \end{gathered}$

Hope this Helps!!!

You might be interested in

What is the slope of the line that passes through (3, −7) and (−1, 1)? (1 point)

ELEN [110]

Answer:

The slope of the line is -2

Step-by-step explanation:

The rule of the slope of a line is:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ , where

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ are two points on the line

Let us find the slope using the rule above

∵ The line passes through points (3, -7) and (-1 , 1)

∴ $x_{1}$ = 3 and $x_{2}$ = -1

∴ $y_{1}$ = -7 and $y_{2}$ = 1

Substitute these values in the rule above to find m

∴ $m=\frac{1--7}{-1-3}=\frac{1+7}{-4}=\frac{8}{-4}=-2$

∴ The slope of the line that passes through the given points is -2

8 0

4 years ago

At what value of x do the graphs of the equations below intersect 2x-y=6 5x+10y=-10

Irina18 [472]

Answer:

The graphs of the equations intersect at x=2

Step-by-step explanation:

we have

$2x-y=6$ ----> equation A

$5x+10y=-10$ ----> equation B

Solve the system by graphing

Remember that the solution of the system of equations is the intersection point both lines

Using a graphing tool

The solution is the point (2,-2)

see the attached figure

therefore

The graphs of the equations intersect at x=2

5 0

3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1