The product of multiplying the ones digit of 59 by 853 is 7677 and the product of multiplying the tens digit of 59 by 853 is 42650 and the final product is 50327
<h3>How to determine the product of the numbers?</h3>
The numbers are given as
853 and 59
By using the standard algorithm i.e. the partial product method, we have the following equation
853 * 59 = 853 * (50 + 9)
Open the bracket
So, we have
853 * 59 = 853 * 50 + 853 * 9
Evaluate the products
So, we have
853 * 59 = 42650 + 7677
The above means that the product of multiplying the ones digit of 59 by 853 is 7677 and the product of multiplying the tens digit of 59 by 853 is 42650
Next, we evaluate the sum
853 * 59 = 50327
This means that the final product is 50327
Read more about products at
brainly.com/question/10873737
#SPJ1
Answer:
C 89a + 2c)
Step-by-step explanation:
The largest factor of both 8 and 16 is 8. So, 8 = GCF
factor out an 8 and you get 8(a + 2c)
To check if you are correct, multiply and see if you get what you started with.
Answer:
Main hu haha
Step-by-step explanation:
hii!! how are you??
Problem 1:
1) 3/5x-1 x 3/4=9/10
3x/5-1 x 3/4=9/10
2) 3x/5-1 x 3/4=9/10
3x/5-3/4=9/10
3) 3x/5-3/4=9/10
+3/4 +3/4
4) 3x/5=33/20
5) 5 x 3x/5= 5 x (33/20)
6) 3x=33/4
7) 3x=33/4
/3 /3
8) x=11/4
Problem 2:
1) 6/7+2/5x=-4/5
6/7+2x/5=-4/5
2) 6/7+2x/5=-4/5
-6/7 -6/7
3) 2x/5= --58/35
4) 5 x 2x/5= 5 x (--58/35)
5) 2x= --58/7
6) 2x= --58/7
/2 /2
x= --29/7