All categories

3 years ago

5 over 8 plus 3 over 10

Mathematics

1 answer:

iogann1982 [59]3 years ago

5 0

Answer:

Step-by-step explanation:

Question: 5/8 + 3/10

Answer: 25/40 + 12/40 = 37/40

Explanation: Find the least common denominator of each fraction

According to this problem, the least common denominator is 40

The common denominator you can calculate as the least common multiple of the both denominators - LCM of 8 and 10 is 40

HOPE I HELPED!!

Please mark as brainliest to help me level up!

You might be interested in

Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution,

Gnom [1K]

Answer:

x1=-4/5, x2=18/5 and x3=7/5

Step-by-step explanation:

$\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right]$

you can do linear combination between the rows:

2nd row=R2-3R1 and 3th row=R3-2R1

$\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right]$

3th row=(3R2-R3)/15

$\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right]$

1st row=R1+3R3 and R2-11R3

$\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right]$

x1=-4/5, x2=18/5 and x3=7/5

6 0

3 years ago

The main show tank has a radius of 60 feet and forms a quarter sphere where the bottom of the pool is spherical and the top of t

svetlana [45]

The given radius is = 60 feet

The volume of the sphere is given by = $\frac{4}{3} \pi r^{3}$

So we will find the volume of the sphere using radius as 60 and pi as 3.14

Volume = $\frac{4}{3}*3.14*60*60*60$ = 902059.20 cubic feet

As quarter of anything is defined as its one-fourth part, so here the volume of the quarter of sphere can be calculated by finding one fourth of the calculated volume.

So volume of quarter sphere = $\frac{902059.20}{4}$

= 225514.80 cubic feet

And rounding this to nearest whole number we get 225515 cubic feet as the answer.

3 0

3 years ago

How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$