MATHEMATICS HIGH SCHOOL

LCD of 3/5x^4 and 2/7x^3

Answers

Answer 1

Answer:

Answer:

31/35x^7

Step-by-step explanation:

3/5x^4 +2/7x^3

you find the lcm of the denominators

21/35x^7 + ,10/35x^7 =31/35x^7

Related Questions

Can someone help me please?1. Refer to the triangle below. If a = 3 and b = 3,A. Find the exact length of c in unsimplified radical formB. Find the exact length of c in simplified radical form C. Find c to the nearest hundredth**I put the pic down below**2. Consider the numbers 4/13 and 3/10 .A. Compare the numbers.B. Find a rational number between the two given numbers.3. Use the Babylonian method to approximate √19 to the nearest hundredth.
a marching band requires a tuba player to begin at the 50 yard line of a football field perpendicular to the 40 yard line. He then makes a 30 degree turn and marches at a diagonal back to the 50 yard line.Once he reaches the 50 yard line he must march forward until; he reaches his original starting position.If the tuba player travels 18 invhes with each marching pace how many paces must he take to complete the routine?
One person sends an email to 2 people. Those 2 people send it to 4 people, and those 4 send it to 8 people. is this linear or exponential?
Write a story problem in which you would have to multiply 5 × (-1/3) to find the answer.
How to solve for 4d-7=25/4

trevon spent 8 hours and 15 minutes at an amusement park yesterday he spent 75% of the time at the park about how much time did he spend on rides

Answers

convert 8 hours and 15 minutes into minutes- 8 x 60=480 480+15=495
100%-75%=25%
25%=0.25
0.25 x 495 = 123.75 minutes or 2 hour and 3.75 minutes

Hope this helps!!

so 8 hours+15mins at park
park=75%
rides=100%-park=100-75=25%
HACK ALERT
25% is 1/4 so multiply 8 hours and 15 mins by 1/4 so
8 h and 15 mins times 1/4=8h times 1/4+15mins times 1/4=2h+3.75min=

2 hours and 3.75 mins spent on rides

How to solve number 21 :/

Answers

$\left(\begin{array}{cc}1&1\n-2&1\end{array}\right) \bold{M}= 2\left(\begin{array}{cc}1&0\n0&1\end{array}\right)\n \bold{M}=2\frac{ \left(\begin{array}{cc}1&0\n0&1\end{array}\right) }{ \left(\begin{array}{cc}1&1\n-2&1\end{array}\right)}\n \bold{M}=2\left(\begin{array}{cc}1&0\n0&1\end{array}\right) \left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)$

$\left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)=(1)/(1-(-2))\left(\begin{array}{cc}1&-1\n2&1\end{array}\right)\n\left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)=(1)/(3)\left(\begin{array}{cc}1&-1\n2&1\end{array}\right)\n\left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)=\left(\begin{array}{cc}(1)/(3)&-(1)/(3)\n(2)/(3)&(1)/(3)\end{array}\right)\n\n$
$\bold{M}= 2\left(\begin{array}{cc}1&0\n0&1\end{array}\right)\left(\begin{array}{cc}(1)/(3)&-(1)/(3)\n(2)/(3)&(1)/(3)\end{array}\right)\n\bold{M}= 2\left(\begin{array}{cc}(1)/(3)&-(1)/(3)\n(2)/(3)&(1)/(3)\end{array}\right)\n\bold{M}= \left(\begin{array}{cc}(2)/(3)&-(2)/(3)\n(4)/(3)&(2)/(3)\end{array}\right)$

Why is 6/8 greater than 5/8 but less than 7/8?