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4 years ago

5 2/3 + 3 1/2 = Your answer

Mathematics

2 answers:

vazorg [7]4 years ago

6 0

Answer:

9 1/6

Step-by-step explanation:

5 2/3 + 3 1/2 =

We need to get a common denominator of 6

5 2/3 *2/2 = 5 4/6

3 1/2 *3/3 = 3 3/6

5 4/6 + 3 3/6

8 7/6

8+ 6/6 + 1/6

8 + 1 + 1/6

9 1/6

Liono4ka [1.6K]4 years ago

4 0

Answer:

9 1/6

Step-by-step explanation:

Add the whole numbers: 5+8

Combine the fractions by making their denominator the same: 7/6

Add 6/6 = 1 so add to the whole numbers to make this a mixed fraction.

You might be interested in

Is the equation a linear function or nonlinear function?

ki77a [65]

Answer:

It is a nonlinear function.

Step-by-step explanation:

All of the numbers and variables actually don't create a straight line, so it is a nonlinear function.

hope this helps.

4 0

3 years ago

Read 2 more answers

Given f(x) =

sergejj [24]

Answer:

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = \left\{ \begin{array}{ll} 2\cos(\pi x) \text{ for } x \leq -1 \\ \\ \displaystyle \frac{2}{\cos(\pi x)}\text{ for } x > -1 \end{array} \right.$

And we want to find:

$\displaystyle \lim_{x\to -1}f(x)$

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

$\displaystyle \lim_{x\to-1^-}f(x)$

Since we are approaching from the left, we will use the first equation:

$\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)$

By direct substitution:

$=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2$

Right-Hand Limit:

$\displaystyle \lim_{x\to -1^+}f(x)$

Since we are approaching from the right, we will use the second equation:

$=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}$

Direct substitution:

$\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2$

So, we can see that:

$\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2$

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

$\displaystyle \lim_{x \to -1}f(x)=-2$

Our answer is A.

8 0

3 years ago

find the median of the following data classes 100-150,150-200,200-250,250-300,300-350 and the frequency is 22,18,25,20,15

kaheart [24]

Answer:

Step-by-step explanation:

median = (N+1)/2

=(100+1)/2

= 50.5

cf just greater than 50.5 is 65 .So, median is 65.

3 0

3 years ago

The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

Mrac [35]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

$A=b^{2}$

where b is the length side of the square

we have

$A=49\ units^2$

substitute

$49=b^{2}$

$b=7\ units$

therefore

$AB=BE=ED=AD=7\ units$

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

$A=(AC)(AG)$

substitute the given values

$A=(9)(10)=90\ units^2$

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

$A=(DE)(DG)$

we have $DE=7\ units$

$DG=AG-AD=9-7=2\ units$

substitute $A=(7)(2)=14\ units^2$

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

$A=(EF)(CF)$

we have

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

substitute

$A=(3)(7)=21\ units^2$

step 5

sum the shaded areas

$14+21=35\ units^2$

step 6

Divide the area of of the shaded region by the area of ACIG

$\frac{35}{90}$

Simplify

Divide by 5 both numerator and denominator

$\frac{7}{18}$

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0

3 years ago

Please ASAP

topjm [15]

Answer:

(a) x +5y = 22

(b) p = 11, q = 5

Step-by-step explanation:

The derivative of a function tells you the slope of its curve at every point. Then the slope of C at x=2 is ...

dy/dx = 2³ +2(2) -7 = 5

The normal to the curve at the point of interest will have a slope that is the opposite reciprocal of this: -1/5. Then the point-slope equation of the normal line can be written as ...

y -k = m(x -h) . . . . line with slope m through point (h, k)

y -4 = -1/5(x -2) . . . . line with slope -1/5 through point (2, 4)

5y -20 = -x +2 . . . . multiply by 5

x +5y = 22 . . . . . . add x+20 to put in standard form

The graph shows curve C and the desired normal line.

The power rule for derivatives tells you ...

(d/dx)(a·x^n) = a·n·x^(n-1)

This relation gives you two ways to find the values of p and q.

a) using the exponent of the term

b) using the coefficient of the term

Either way, the values are ...

p = 10 +1 = 11

q = 4 +1 = 10/2 = 5

6 0

2 years ago