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3 years ago

Find the midpoint of points A(-7,-6) and B(1,-10) graphically.

Mathematics

1 answer:

mixas84 [53]3 years ago

4 0

-3, -8 i think lmk .

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I need help on this During the spring, Nina Milling assembles bicycles at The Wheeler Dealer. She is paid $12.00 for each bicycl

frez [133]

Answer:

$556

Step-by-step explanation:

4+7+6+10+8=35

35x12=420

4x14=56

5x16=80

420+56+80=556

7 0

3 years ago

Read 2 more answers

The point(5/13,Y) IS IN THE 4TH quadrant corresponds to angle on the unit circle

shutvik [7]

4th quadrant is positive x (cos) and negative y (sin)

Use the Pythagorean Theorem to calculate the value of y (sin).
x² + y² = c²
5² + y² = 13²
25 + y² = 169
y² = 144
y = 12

Since the y-value is negative in the 4th quadrant, then y = -12

6 0

3 years ago

Solve for x <br> (1 point)

Delicious77 [7]

It should be 13 since 13+8 = 21. 21-7 = 13

5 0

3 years ago

What math operation determines the pattern? 48, 3, 45, -42?

34kurt

Just subtract the following number from the preceding one.
48- 3= 45
3- 45= -42.

8 0

3 years ago

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

1 year ago

Find the midpoint of points A(-7,-6) and B(1,-10) graphically.​

Find the midpoint of points A(-7,-6) and B(1,-10) graphically.