Let's factor all four of these before we multiply.
x² - 25 We want two numbers that add to -25 and multiply to 0.
= (x+5)(x-5)
x² - 2x
Both terms are divisible by x.
= x(x-2)
x - 2unfactorable
x² + 5x
Both terms are divisible by x.
= x(x+5)
Now we have this:
Let's go ahead and multiply across.
Cancel out the (x+5) and (x-2) from the top and the bottom.
Answer:
The minimum number of letters John has to send to be sure that Peter receives his letter is 127 letters
Step-by-step explanation:
The four digit numbers that are multiples of 5 and 7 with the last digit = 0 is found as follows
Since the last digit of the house number = 10, then the house number is divisible by 10 which also meets the condition that the house number is divisible by 5
We have the four digit numbers from 1000 to 9999
Hence the numbers divisible by both 7 and 10 are from (1000/70 (Which is 14 + 2/7) - 2/7)×70 + 70 = 1050 to (9999/70 (Which is 142 + 59/70)- 59/70)×70= 9940
Which gives 142 - 15 = 127 numbers which are four digit number multiples of 5 and 7 with the last digit = 0
Hence the minimum number of letters John has to send to be sure that Peter receives his letter = 127 letters.
-45.6 is the answer if you evaluate
Answer:
Pythagorean theorem which is a²+b²=c2
9²+12²=c²
225=c2
c=√225
c=15
Answer:
Step-by-step explanation:
![\sf x(4+5) = 54\\\\Applying \ distributive \ property\\\\(x*4) + (x*5) = 54\\\\Distributive\ Property\ is:\\\\A(B+C) = (A*B)+(A*C)\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%20x%284%2B5%29%20%3D%2054%5C%5C%5C%5CApplying%20%5C%20distributive%20%5C%20property%5C%5C%5C%5C%28x%2A4%29%20%2B%20%28x%2A5%29%20%3D%2054%5C%5C%5C%5CDistributive%5C%20Property%5C%20is%3A%5C%5C%5C%5CA%28B%2BC%29%20%3D%20%28A%2AB%29%2B%28A%2AC%29%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
