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tino4ka555 [31]
3 years ago
12

Find two numbers whose sum is 15 and the product is 36

Mathematics
2 answers:
ycow [4]3 years ago
8 0
The two numbers are 3 and 12.

I got these numbers by factoring 36.

1 * 36    sum : 37
2 * 18    sum : 20
3 * 12    sum : 15

It is easier to find the factor of a number and solve for the sum. Than to find the addends of a number and solve for the product.

There are 9 factors of 36 while 14 addends of 15.
Vedmedyk [2.9K]3 years ago
7 0
OK so if you're wanting 15 and 36 you will need to know what goes into each of them but if you just need 15 then let's try to find that one so, what does go into 15 well let's find out.

15:1×15,3×5
36:1×36,2×18,3×12,4×9,6²

so, based on what we have here there are a lot of answers between numbers 15&36 but one maybe two or more answers for what can go into 15 and 36.

3×12=36
12+3=15
so, the two numbers you need are 3 and 12.

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50 w +100 is greater than or equal to 18000
Marrrta [24]
What you would do to “simplify” the equation is 18000-100 then divide by 50. The answer would be 358. Hoped this helped!
8 0
3 years ago
Find the missing side of this right
I am Lyosha [343]

Answer:

Step-by-step explanation:

square of 15 + square of x = square of 21

225 + square of x = 441

441 - 225 = 216

x = 216

6 0
3 years ago
Type the correct answer in each box. (____)
trapecia [35]

Answer:

1. -16; 2. +64; 3. 16

Step-by-step explanation:

The formula for the volume of a cylinder is

V = πr²x

Data:

r = (x - 8)   mm

V = 1024π mm³

Calculations:

1. Find the cubic equation  

V = πr²h

1024π = π(x - 8)² × x

Divide each side by π

1024 = x(x - 8)²  

1024 = x(x² - 16x  + 64)

1024 =    x³ - 16x² + 64x

x³ - 16x² + 64x - 1024 = 0

2. Solve the cubic equation

The general formula for a third-degree polynomial is

f(x) = ax³ + bx² + cx + d

Your polynomial is  

f(x) = x³ - 16x² + 64x - 1024

a = 1; d = -1024

According to the <em>rational roots theorem</em>, the possible roots are

factors of d/factors of a

Factors of d = ±1, ±2, ±4, ±8, ±16, ± 32, ± 64, ±128, ±256, ±512, ±1024

Factors of a = ±1

Potential roots are x = ±1, ±2, ±4, ±8, ±16, ± 32, ± 64, ±128, ±256, ±512, ±1024

That's a lot of possibilities to check by trial and error. I will just use the one that works.

Try x = 16 by synthetic division.

16|1  -16   64  -1024

  <u>|     16     0   1024 </u>

   1     0   64        0

So,

(x³ - 16x² + 64x - 1024)/(x – 16) = x² + 64

and

(x - 16)(x² + 64) = 0

x - 16 = 0        x² + 64 =    0

     x = 16       x²         = -64

                       x          =  ±8i

There is only one real root: x = 16 mm

6 0
3 years ago
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4 0
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