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Contact [7]
3 years ago
10

What two numbers add up to 10 and multiply to 21

Mathematics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:The answer is 3 & 7

Step-by-step explanation:3 x 7 = 21

3 + (7) = 10

Are you asking because you are trying to figure out how to factor the following quadratic equation?

x2 + 10x + 21 = 0

If so, the solution to factor the quadratic equation above is:

(X + 3 ) (X + 7)

To summarize, since 3 and 7 multiply to 21 and add up 10, you know that the following is true:

x2 + 10x + 21 = (X + 3 ) (X + 7)

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What is the sum of 12 – 5i and –3 + 4i?
8090 [49]
12 - 3 =9 and -5i+4i=1i or i so it is      {9-i}
6 0
3 years ago
Read 2 more answers
How do the constant of proportionality and slope relate to the unit rate
Pepsi [2]

Answer:

When a relationship is proportional, all y over x ratios simplify to the slope of the line which is the constant of proportionality or the unit rate, with one as a denominator. Example: Consider the equation y = 60x. The slope is 60. The rate of change is 60 units for every one unit.

Step-by-step explanation:

7 0
3 years ago
-7x+12-2x=23+13x simplify by combining like terms that are on the same side of the equation
zavuch27 [327]
-7x+12-2x=23+13x

First take away 13x from the right side and put it on the left side adding taking it away from -7x 

-7x(-13x)=-20x
-20x+12-2x=23

then add 2x to -20x
and move 12 to the other side, minus it off of 23

-18x=11

then divide -18x on both sides 

then x=-11/18

6 0
4 years ago
A. 90<br><br> B. 105 <br><br> C. 225<br><br> D. 315
Simora [160]

Answer:

225 is the correct one

Step-by-step explanation:

u multiply 15m*9 months u get 135 and then u add the existing 90m and then it becomes 225mp3

15m*9+90

7 0
3 years ago
Suppose that P(n) is a propositional function. Determine for which nonnegative integers n the statement P(n) must be true if a)
wel

Solution :

a). $P(0)$ is true

Then ,$P(0+2)=P(2)$ is true.

         $P(2+2)=P(4)$ is true

          $P(4+2)=P(6)$ is true.

Therefore, we see that $P(n)$ is true for all the even integers : $\{0, 2,4,6,...\}$

b). $P(0)$ is true

Then ,$P(0+3)=P(3)$ is true.

         $P(3+3)=P(6)$ is true

          $P(6+3)=P(9)$ is true.

Therefore, we see that $P(n)$ is true for all the multiples of 3 : $\{0, 3,6,9,12,...\}$

c). $P(0)$ and $P(1)$ is true, then $P(0+2)=P(2)$ is true

$P(1)$ and $P(2)$ is true, then $P(1+2)=P(3)$ is true.

$P(2)$ and $P(3)$ is true, then $P(2+2)=P(4)$ is true.

So, we observe that  $P(n)$ is true for all the non- negative integers : $\{0, 1,2,3,4,5,6,...\}$.

d). $P(0)$ is true,

   So, $P(0+2)$ and $P(0+3)$ is true or $P(2)$ and $P(3)$ is true.

   Now,   $P(2)$ is true.

Again, $P(2+2)$ and $P(2+3)$ is true or $P(4)$ and $P(5)$ is true.

   Now, $P(3)$ is true.

Again, $P(3+2)$ and $P(3+3)$ is true or $P(5)$ and $P(6)$ is true.

Thus,

$P(n)$ is true for all the non- negative integers except 1 : $\{0, 2,3,4,5,6,...\}$.

3 0
3 years ago
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