Ask question

Ask question

All categories

3 years ago

6

Please put the fraction in lowest terms 26/30 i am her little bro btw my sis said u guys would some how help

2 answers:

avanturin [10]3 years ago

5 0

Divide both the numerator and denominator by the highest number that can go into both.
26 divide 2 13
----- -----
30 divide 2 15

ans : 13/15 . Whatever is the highest number that can evenly go into both the top and bottom is the number you divide by until you cannot divide it

aniked [119]3 years ago

5 0

Divide 26 by 2 to get 13. Divide 30 by 2 to get 15. This cannot be simplified any more so your answer is 13/15. Happy to help. :)

You might be interested in

Write and solve a proportion that the teacher can use to estimate how many students in the whole school would choose the aquariu

Anon25 [30]

Answer:

Never gonna

Step-by-step explanation:

give you up,

Let you down,

make you cry

Run around and desert you

6 0

2 years ago

Help with my math i will mark brainliest

Svetlanka [38]

Answer:

D

Step-by-step explanation:

step 1: find the slope

slope is 2/-4 = -1/2

then plug in the value for x to y, you get +1 for the y intercept

then it becomes y=-1/2x + 1

I hope this answer helped, and if it was correct, please mark this as brainliest! Thank you!

4 0

2 years ago

Read 2 more answers

Describe importance of public participant

Effectus [21]

Answer:

the main aim of public participant is to encourage the public to have a meaningful input into the decision making process.

6 0

3 years ago

Consider the given matrix. 1 −1 0 5 1 4 0 1 1 Find the eigenvalues. (Enter your answers as a comma-separated list.) λ = Find the

fomenos

Answer:

Step-by-step explanation:

We have to diagonalize the matrix

$\left[\begin{array}{ccc}1&-1&0\\5&1&4\\0&1&1\end{array}\right]$

we have to solve the expression

$|A-\lambda I|=0$

Thus, by applying the determinant we obtain the polynomial

$(1-\lambda )^3+5-4=0\\(1-\lambda )^3+1=0\\$

$-\lambda^3+3\lambda^2-4\lambda+2$

$\lambda_1=1\\\lambda_2=1-i\\\lambda_3=1+i\\$

and the eigenvector will be

$v_1=(-4,0,5)\\v_2=(-1,-i,-1)\\v_3=(-1,i,1)$

HOPE THIS HELPS!!

3 0

3 years ago

Consider the following set of equations, where s, s0, x and r have units of length, t has units of time, v has units of velocity

Anna11 [10]

Answer:

5

Step-by-step explanation:

For an equation to be dimensionally correct the dimension of quantities on both sides of equation must be same.

Also, two physically quantities can only be added or subtracted only when their dimension are same.

here all option are dimensionally correct except the 5th option where

dimension of t= [T] whereas dimension of a/v is $\frac{LT^{-2}}{LT^{-1}}$

= T^{-1}

since, the dimension of quantities on either sides of equation are not the same the equation is dimensionally is incorrect.

8 0

2 years ago