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PSYCHO15rus [73]

3 years ago

12

I need help !!!!!!!!!

1 answer:

german3 years ago

5 0

Answer:

a. 14

b. 10

c. 11

Step-by-step explanation:

You need to use the distance formula.

$\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2}}$

Ex: (1,2) and (1,16)

You will need to lable each

1st Point: 1 will be $x_{1}$ and 2 will be $y_{1}$

2nd Point: 1 will be $x_{2}$ and 16 will be $y_{2}$

Then you will need to plug it in

$\sqrt{(1-1)^{2} +(16-2)^{2}}$

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For 902798 write the digit in the ten thousand place

netineya [11]

The digit in the ten thousand place is
0

7 0

4 years ago

I could use alot of help on this please and thankyouuuu

kicyunya [14]

Answer:

B. Step 4 is incorrect , it should be $\frac{10}{8}$

Step-by-step explanation:

6x-1= -2x+9

Step 1

Collect like terms and change signs if you move anything from one side to the other; 6x+2x = 9+1

Step 2: Add like terms on both sides

8x = 10

Step 3: To remain with just x on one side, divide both sides by 8 and it becomes;

$\frac{8x}{8} = \frac{10}{8}$

The two 8s cancel each other and you remain with $x=\frac{10}{8}$

6 0

4 years ago

Find the inverse of square root f(n) = n – 3.

pogonyaev

Answer:

$f'(n) = n + 3$

Step-by-step explanation:

Given

$f(n) = n - 3$

Required

The inverse

$f(n) = n - 3$

Replace f(n) with y

$y = n - 3$

Swap positions of y and n

$n = y - 3$

Make y the subject

$y = n + 3$

Replace y with f'(n)

$f'(n) = n + 3$

4 0

3 years ago

19) Albert says that the two systems of equations shown have the same solutions.

k0ka [10]

Answer:

option A) Agree, because the solutions are the same is correct.

Step-by-step explanation:

FIRST SYSTEM

$6x + y= 2$

$-x-y=-3$

solving the system

$\begin{bmatrix}6x+y=2\\ -x-y=-3\end{bmatrix}$

$\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}6\:\mathrm{:}\:\quad \:-6x-6y=-18$

$\begin{bmatrix}6x+y=2\\ -6x-6y=-18\end{bmatrix}$

adding the equation

$-6x-6y=-18$

$+$

$\underline{6x+y=2}$

$-5y=-16$

so the system becomes

$\begin{bmatrix}6x+y=2\\ -5y=-16\end{bmatrix}$

solve -5y for y

$-5y=-16$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-16}{-5}$

simplify

$y=\frac{16}{5}$

$\mathrm{For\:}6x+y=2\mathrm{\:plug\:in\:}y=\frac{16}{5}$

$6x+\frac{16}{5}=2$

subtract 16/5 from both sides

$6x+\frac{16}{5}-\frac{16}{5}=2-\frac{16}{5}$

$6x=-\frac{6}{5}$

Divide both sides by 6

$\frac{6x}{6}=\frac{-\frac{6}{5}}{6}$

$x=-\frac{1}{5}$

Therefore, the solution to the FIRST SYSTEM is:

$x=-\frac{1}{5},\:y=\frac{16}{5}$

SECOND SYSTEM

$2x-3y = -10$

$-x-y=-3$

solving the system

$\begin{bmatrix}2x-3y=-10\\ -x-y=-3\end{bmatrix}$

$\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y=-6$

$\begin{bmatrix}2x-3y=-10\\ -2x-2y=-6\end{bmatrix}$

$-2x-2y=-6$

$+$

$\underline{2x-3y=-10}$

$-5y=-16$

so the system of equations becomes

$\begin{bmatrix}2x-3y=-10\\ -5y=-16\end{bmatrix}$

solve -5y for y

$-5y=-16$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-16}{-5}$

Simplify

$y=\frac{16}{5}$

$\mathrm{For\:}2x-3y=-10\mathrm{\:plug\:in\:}y=\frac{16}{5}$

$2x-3\cdot \frac{16}{5}=-10$

$2x=-\frac{2}{5}$

Divide both sides by 2

$\frac{2x}{2}=\frac{-\frac{2}{5}}{2}$

Simplify

$x=-\frac{1}{5}$

Therefore, the solution to the SECOND SYSTEM is:

$x=-\frac{1}{5},\:y=\frac{16}{5}$

Conclusion:

As both systems of equations have the same solution.

Therefore, we conclude that Albert is right when says that the two systems of equations shown have the same solutions.

Hence, option A) Agree, because the solutions are the same is correct.

8 0

3 years ago

No need to answer, just the points!!<br>

Ipatiy [6.2K]

Answer:

WAIT WHAT

Step-by-step explanation:

7 0

3 years ago