• There is a 7 centimeter difference
• They both have 10 centimeters
• They are both made out of string
• They both use centimeters
Answer:
<u>R8000</u>.
Step-by-step explanation:
First, let's calculate the amount they are going to have to pay the bank after 1 year. It is given by the following expression:
R5 000 * 1.28= R6400
Therefore, they have to pay R6400/12 monthly.
Now, in fifteen months the total amount to pay should be (R6400/12) * 15= <u>R8000</u>.
Sum of three numbers is 33 and sum of their cubes is 5049 then three numbers are 7,11,15 or 15,11,7.
As given,
Let a-d, a, a +d be three numbers
Sum =33
⇒a-d + a +a +d =33
⇒ a =11
Sum of cubes= 5049
⇒ (a-d)³ + a +(a +d)³ =5049
⇒ a³ -d³ -3a²d +3ad² + a³ +a³ +d³ +3a²d +3ad² =5049
⇒3a³ + 6ad² = 5049
⇒ d=± 4
Therefore, sum of three numbers is 33 and sum of their cubes is 5049 then three numbers are 7,11,15 or 15,11,7.
Learn more about numbers here
brainly.com/question/17429689
#SPJ4
Answer:
x = 
Step-by-step explanation:
Given
f - ex = g - hx
Collect the terms in x on the left side by adding hx to both sides
hx - ex + f = g ( subtract f from both sides )
hx - ex = g - f ← factor out x on the left side
x(h - e) = g - f ← divide both sides by (h - e)
x =
Answer:
Step-by-step explanation:
11) First write in decreasing exponential terms and fill in blanks with zeros.
Our goal is to eliminate all term in the dividend by subtraction
________________
2v - 2 | 2v³ - 16v² + 0v + 13
we see that 2v needs to be multiplied by 1v² to eliminate the first term
<u> v² </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v²
multiply your estimate by your divisor and subtract from the dividend.
bring down the next term and repeat.
<u> v² -7v </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v² + 0v
<u>-(-14v² + 14v)</u>
- 14v
repeat again
<u />
<u> v² - 7v - 7 </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v² + 0v
<u>-(-14v² + 14v)</u>
- 14v + 13
<u>-(-14v + 14)</u>
-1
and remainder gets put over the divisor and appended
v² - 7v - 7 - 1/(2v - 2)
13) Same process
<u> </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u> 8a² </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
<u> 8a² - 4a </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
-(-<u>20a² - 4a)</u>
-35a - 5
<u> 8a² - 4a - 7 </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
-(<u>20a² - 4a)</u>
-35a - 5
-<u>(-35a - 7)</u>
2
8a² - 4a - 7 + 2/(5a + 1)
