All categories

3 years ago

Which is equivalent to 2 5/6A. 5/3B. 7/6C. 17/6D. 12/5

Mathematics

2 answers:

Dimas [21]3 years ago

8 0

The answer would be C (17/6)
This is because;
6 goes into 17 twice leaving a reminder of 5.

RSB [31]3 years ago

8 0

Convert 2 5/6 into an improper fraction by multiplying 6 and 2 then adding 5. Keep the same denominator. The improper fraction would be 17/6. So your answer would be, C, 17/6.

You might be interested in

Please Help!! Mathematics Grade 8

Evgen [1.6K]

you need to get a common denominator and subtract 20/32 from 26/32
z would equal 6

4 0

3 years ago

WILL GIVE BRAINLIST!!! ANSWER QUICK

slavikrds [6]

Answer:Y=3x(x-4)

Step-by-step explanation:

4 0

2 years ago

The sum of two numbers is 39 and the difference is 3. What are the numbers?

frosja888 [35]

Answer: 21 & 18

Step-by-step explanation:

Sum of two numbers is 39

Difference is 3

Let the numbers be x & y

X+y=39........equation 1

X-y=3...........equation 2

X=3+y............equation 3

Substitute equation 3 into equation 1

(3+y)+y=39

3+y+y=39

3+2y=39

2y=39-3

2y=36

Y=36/2

Y=18

Substitute for y in equation 2

X-y=3

X-18=3

X=3+18

X=21

The two numbers are 21 & 18

8 0

3 years ago

The area of a parrellogram is 31.5ft and height is 6.3ft. What is the length of the base

Musya8 [376]

Answer:

Area = base × height

31.5 = base × 6.3

base = 5 ft

8 0

3 years ago

Read 2 more answers

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago