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sattari [20]
2 years ago
5

I need help ty-math-

Mathematics
2 answers:
Dmitry_Shevchenko [17]2 years ago
7 0

Answer:

1,2,2,2,2,2,2,3,3,4,5,5,6

Step-by-step explanation:

Anastasy [175]2 years ago
5 0

<em>Oh, math! One of my absolute favorite subjects. I'd be happy to help.</em>

<em>From least to greatest, here's the list-</em>

<em>1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 10 </em>

<em />

<em>I hope this answers your question.</em>

<em>-Toremi</em>

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romanna [79]
You have to factor out the 3
7 0
3 years ago
Twenty words were randomly selected from two textbooks. Each set gives the number of letters in each of the twenty words for tha
Alborosie

Answer:

k = 13The smallest zero or root is x = -10

Step-by-step explanation:

you can write "x^2" to mean "x squared"

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

7 0
3 years ago
What is the range of the data below?<br>100<br>105<br>110<br>115<br>120<br>125<br>A 2 B 5 C 12 D 13​
MrMuchimi

Answer:

ifk

Step-by-step explanation:

8 0
3 years ago
Can anyone please help me in this
Alik [6]

These problems are called systems of equations. Basically you have two linear equations and you need to find the values for x and y. In other words, all these equation are lines and our answer will be the exact point that the pair of lines intersect. For example, if we get x=1 and y=2 the lines will intersect at point (1,2). Now that you have some background knowledge here comes the tricks and tactics kid.

We know that we can solve one variable equation easily. For example...

x+1=2

x=1 obviously

Cause we have two variables x and y it is not possible to find a solution. For example, in the equation x+y=10, x=1 when y=9 and x=2 when y=8. There is not correct answer.

So what can we do? We have to make a two variable equation into a one variable equation.

There are two ways to do this: substitution and elimination. I will create a sample problem and then solve it using both methods.

x+y=2

2y-y=1

3)  

-3x-5y=-7 -----> -12x-20y=-28

-4x-3y=-2 ------> -12x-9y=-6

 -12x-20y=-28

-(-12x-9y=-6)

---------------------

-11y=-22

y=2

-3x-5(2)=-7

-3x=3

x=-1

4) 8x+4y=12 ---> 24x+12y=36

7x+3y=10 ---> 28x+12y=40

 28x+12y=40

-(24x+12y=36)

---------------------

4x=4

x=1

8(1)+4y=12

4y=4

y=1

5) 4x+3y=-7

-2x-5y=7 ----> -4x-10y=14

    4x+3y=-7

+(-4x-10y=14)

-------------------

-7y=7

y=-1

4x+3(-1)=-7

4x=-4

x=-1

6) 8x-3y=-9 ---> 32x-12y=-36

5x+4y=12 ---> 15x+12y=36

 32x-12y=-36

+(15x+12y=36)

--------------------

47x=0

x=0

8(0)-3y=-9

-3y=-9

y=3

7)-3x+5y=-2

2x-2y=1 ---> x-y=1/2 ----> x=y+1/2

-3(y+1/2)+5y=-2

-3y-1.5+5y=-2

2y=-0.5

y=0.25

2x-2(0.25)=1

2x=1.5

x=0.75

4 0
3 years ago
Number 1 and 2 please (25 points)
Elenna [48]

Answers:

1) (3x+2)+(4-5x)=-2x+6

(3+2i)+(4-5i)=7-3i

2) (1-3x)(2+x)=-3x^{2}-5x+2

(1-3i)+(2+i)=5-5i

Step-by-step explanation:

In mathematics there are rules related to complex numbers, specifically in the case of addition and multiplication:

<u>Addition: </u>

If we have two complex numbers written in their binomial form, the sum of both will be a complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imaginary parts (similarly as the sum of two binomials).

For example,  the addition of these two binomials is:

(3x+2)+(4-5x)=3x-5x+2+4=-2x+6

Similarly, the addition of two complex numbers is:

(3+2i)+(4-5i)=3+4+2i-5i=7-3i Here the complex part is the number with the i

<u>Multiplication: </u>

If we have two complex numbers written in their binomial form, the multiplication of both will be the same as the multiplication (product) of two binomials, taking into account that i^{2}=-1.

For example,  the multiplication of these two binomials is:

(1-3x)(2+x)=2+x-6x-3x^{2}=-3x^{2}-5x+2

Similarly, the multiplication of two complex numbers is:

(1-3i)+(2+i)=2+i-6i-3i^{2}=2-5i-3(-1)=5-5i

8 0
3 years ago
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