Can someone plz help me with this one problem plzzzzz!!!!

Plz help

Yuliya22 [10]

Answer:

Solution: x = 2, y = -1 or (2, -1)

Step-by-step explanation:

Equation 1: 2x + y = 3

Equation 2: 5x - 2y = 12

Using the substitution method:

Transform the Equation 1 into its slope-intercept form:

2x + y = 3

2x - 2x + y = -2x + 3

y = 2x + 3

Substitute the value of y = -2x + 3 into Equation 2:

5x - 2y = 12

5x - 2(-2x + 3) = 12

5x + 4x - 6 = 12

9x - 6 = 12

9x - 6 + 6 = 12 + 6

9x = 18

9x/9 = 18/9

x = 2

Substitute the value of x = 2 into Equation 2 to solve for y:

5x - 2y = 12

5(2) - 2y = 12

10 - 2y = 12

10 - 10 - 2y = 12 - 10

-2y = 2

-2y/-2 = 2/-2

y = -1

Double-check whether the values for x and y will provide a true statement for both equations:

Equation 1: 2x + y = 3

2(2) + (-1) = 3

4 - 1 = 3

3 = 3 (True statement)

Equation 2: 5x - 2y = 12

5(2) - 2(-1) = 12

10 + 2 = 12

12 = 12 (True statement)

Therefore, the correct answers are: x = 2; y = -1 or (2, -1).

4 0

3 years ago

Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

Step-by-step explanation:

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

3 0

3 years ago

The set of x valúes of a function this is another name of range

marin [14]

Answer:

False?

Step-by-step explanation:

We can't tell what the question is here, but it looks like you want to know about the definition of domain and range.

The set of x-values is the domain.

The set of y-values is the range.

__

"range" is not another name for the set of x-values.

8 0

3 years ago

1.What are the zeros of the polynomial function?

Lorico [155]

Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________

To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9

Well, this would be your answers.

7 0

3 years ago

Callie surveyed 40 students at her school to see how many have dogs and/or cats. She found that 40% of students who have dogs, d

irina [24]

I think you need to show the tablle

6 0

3 years ago